COUNTING

BASICS
14 NOVEMBER 2023
REVISION: 359

AZIZ MANVA
AZIZMANVA@GMAIL.COM

ALL RIGHTS RESERVED

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TABLE OF CONTENTS
3.2 Applications of Lists 56
TABLE OF CONTENTS ................................. 2 3.3 Visualizing Sets 65
1. SET THEORY............................................ 3 3.4 Venn Diagrams: Two Sets 76
3.5 Venn Diagrams: Three Sets 90
1.1 Set Notation and Properties 3
1.2 Classifying Numbers 8 4. COUNTING RULES ................................ 99
1.3 Operations on Sets 14
4.1 Addition and Multiplication Rules 99
2. COUNTING STRATEGIES ..................... 21 4.2 Using the Rules 105
4.3 Repetition 115
2.1 Enumeration 21
4.4 Counting Numbers and Words-I 125
2.2 Geometrical Counting 26
4.5 Difficult Questions on Counting 134
2.3 Complementary Counting 36
4.6 Preparing for Probability 147
2.4 Reframing the Question / Clever Logic 37
4.7 Further Applications of Counting 169
3. COUNTING WITH SETS ....................... 40 4.8 Sets, Relations and Functions 173

3.1 Single Set: Lists 40

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1. SET THEORY
1.1 Set Notation and Properties
A. Definition and Basics

1.1: Definition of Sets

A set is a collection of well-defined objects.

Example 1.2
Decide which of the following is a set:
A. Tall girls in Mumbai.
B. Girls in Mumbai taller than 5 feet 2 inches.
C. Expensive restaurants in Mumbai.

𝐴 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑠𝑒𝑡 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑡𝑎𝑙𝑙 𝑐𝑎𝑛 𝑏𝑒 𝑝𝑒𝑟𝑐𝑒𝑖𝑣𝑒𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑙𝑦.

𝐵 𝑖𝑠 𝑎 𝑠𝑒𝑡.

1.3: Elements of a set

The objects which make up a set are called its elements, or also its members.

Example 1.4
What are the elements of the set 𝐴 = {𝑎, 𝑏, 𝑐}

There are three elements, which are 𝑎, 𝑏 and 𝑐.

1.5: Roster Notation

In roster notation, we write the elements of the set within curly braces.

Example 1.6
Write the set with elements 1, 2, and 3 in roster notation. Call the set B.

𝐵 = {1, 2, 3}

Example 1.7
𝑥 is an integer that is greater than −3 and less than 4. Let 𝑋 represent the set of values that 𝑥 can take. Write 𝑋
in roster notation.

𝑋 = {−2, −1,0,1,2,3}

1.8: Ellipsis
Three dots are used to indicate continuation to indicate continuation of a series.
has ten elements in it – the ten numbers from 1 to 10.

Dots lets us write larger sets without writing them all out.

Example 1.9
Write {1,2,3,4,5,6,7,8,9,10} using ellipsis.

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{1, 2, … , 10}

Example 1.10
Write the set of natural numbers from 1 to 500 using roster notation.

Writing them all out would be very time-consuming. We use ellipsis:

{1, 2, … ,500}
B. Properties of Sets

1.11: Equal Sets

If two sets have the same elements, they are called equal sets.

Example 1.12
Are the sets 𝐻 = {𝐴, 𝐵, 𝐶} and 𝑍 = {𝐴, 𝐵, 𝐶} equal?

Yes

1.13: Repetition of Elements

Elements are not repeated in a set – they are written only once.

Example 1.14
Are the sets {1, 2, 3, 2} and {1, 2, 3} equal?

Yes

1.15: Order
Order in which elements are written is not important in a set.

Example 1.16
Are the sets {1, 2, 3} and {1, 3, 2} equal?

Since order is not important in a set, the two sets are equal:
{1, 2, 3} = {1, 3, 2}

1.17: Set Builder Notation

Here, a rule to decide the members of the set is written after a colon.

Example 1.18
Convert {1, 2, 3} into set builder notation.

𝐴 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 < 4}

Example 1.19
List the sets below in roster form:
A. {𝑥: 𝑥 is a natural number ≤ 4}
B. {𝑥: 𝑥 is a natural number > 8}

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C. {𝑥: 𝑥 is a natural number ≥ 8}

{1, 2, 3, 4}
{9, 10, 11, … }
8, 9, 10, … }
C. Special Sets
Some sets are important enough for them to have names.

1.20: Universal Set

The set of all elements under consideration is called the Universal set.

1.21: Null Set / Empty Set

A set with no elements is called a null set.

𝜙, { } are ways of writing a null set

𝜙 is the Greek Letter 𝑃ℎ𝑖.

Note: This is different from the Greek Letter 𝜋, which you might have already seen in Geometry.

1.22: Singleton Set

A set with only one element in it is called a singleton set.

Examples
A = {3}
H = {𝑥: 𝑥 is a prime number between 50 and 58}

1.23: Finite and Infinite Sets

A set with a finite number of elements is called a finite set.
A set with an infinite number of elements is called an infinite set.

1015 = 1 𝑞𝑢𝑎𝑑𝑟𝑖𝑙𝑙𝑖𝑜𝑛
1018 = 1 𝑞𝑢𝑖𝑛𝑡𝑖𝑙𝑙𝑖𝑜𝑛

Example 1.24
Classify the following sets as finite or infinite.
A. H = {𝑥: 𝑥 is a prime number between 50 and 100}
B. H = {𝑥: 𝑥 is a prime number}

1.25: Disjoint or Mutually Exclusive Sets

Are sets which have no common elements

It is important to distinguish between a large (but finite) set, and an infinite set.

Infinity (Advanced)
∞
The concept of infinity is related to sets. There are many sets with infinite elements.
Infinity does not follow the usual rules.

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For example:
Addition
3+2=5 X+2=X+2 ∞+2=∞

D. Cardinality

1.26: Cardinality
Cardinality is the number of elements of a set
Cardinality is written by using a small 𝑛 before the name of the set.

Example 1.27
What is the cardinality of the set 𝐴 = {1, 2, 3}?

𝑛(𝐴) = 𝑛{1,2,3} = 3

Example 1.28
Find the cardinality of each set.
A. H = {1, 2, …,12, 13}
B. I = {12, 13, . . . 27, 28}
C. J = {32, 34, . . . 76, 78}
D. K = {57, 59, . . . 113, 115}

A. H = {1, 2, . . . 12, 13} has cardinality 13.

n(H) = 13

{12, 13, . . . 27, 28} = {𝟏, 𝟐, … , 𝟏𝟏, 12, 13, . . . 27, 28} = 28 − 11 = 17
B. n(I) = 28 – 12 + 1 = 17
C. n(J)
= Cardinality of {16, 17, . . . 38, 39}
= 39 – 16 + 1 = 24
D. n(K)
= Cardinality of {56, 58, . . . 112, 114}
= Cardinality of {28, 29, . . . 56, 57}
= 57 – 28 + 1 = 30

E. Belongs To

1.29: Belongs To
The symbol
∈
Which means “belongs to” is used to indicate that an element is a member of a set.

1.30: Does Not Belong To

The symbol
∉
Which means “does not belong to” is used to indicate that an element is not a member of a set.

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Example 1.31
𝑋 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑡𝑤𝑜 𝑑𝑖𝑔𝑖𝑡 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 }
𝑃 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑝𝑟𝑖𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟}

Decide if each statement below is true or false:

A. 4 ∈ 𝑃
B. 2 ∈ 𝑃
C. 1 ∈ 𝑃
D. 7 ∈ 𝑋
E. 9 ∉ 𝑃
F. 5 ∉ 𝑋

Type equation here.

F. Subsets and Supersets

1.32: Subsets
If every element of the set 𝑋 is also an element of the set 𝑌, then 𝑋 is a subset of 𝑌.
𝑋 𝑖𝑠 𝑎 𝑠𝑢𝑏𝑠𝑒𝑡 𝑜𝑓 𝑌 ⇔ 𝑋 ⊆ 𝑌

1.33: Proper Subsets

If 𝑋 is a subset of 𝑌, and there is at least one element in 𝑌 which is not in 𝑋, then 𝑋 is a proper subset of 𝑌.
𝑋 𝑖𝑠 𝑎 𝑝𝑟𝑜𝑝𝑒𝑟 𝑠𝑢𝑏𝑠𝑒𝑡 𝑜𝑓 𝑌 ⇔ 𝑋 ⊂ 𝑌

1.34: Superset
If every element of the set 𝑋 is also an element of the set 𝑌, then 𝑌 is a superset of 𝑋.
𝑌 𝑖𝑠 𝑎 𝑠𝑢𝑝𝑒𝑟𝑠𝑒𝑡 𝑜𝑓 𝑋 ⇔ 𝑌 ⊇ 𝑋

1.35: Proper Superset

If 𝑌 is a superset of 𝑋, and there is at least one element in 𝑌 which is not in 𝑋, then 𝑌 is a proper superset of 𝑋.
𝑌 𝑖𝑠 𝑎 𝑠𝑢𝑝𝑒𝑟𝑠𝑒𝑡 𝑜𝑓 𝑋 ⇔ 𝑌 ⊃ 𝑋

Example 1.36
Consider the sets:
𝑈 = 𝑈𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙 𝑆𝑒𝑡 = {𝑋: 1 ≤ 𝑋 ≤ 10, 𝑋 ∈ ℕ}
𝐴 = {1,3,5}
𝐵 = {2,3,5}
𝐶 = {5,3,1}
𝐷 = {1,2,3,5}

A. Is 𝐴 a subset of 𝐷?
B. Is 𝐷 a subset of 𝐴?
C. Is 𝐴 a subset of 𝐶?
D. Is 𝐶 a subset of 𝐴?
E. Is 𝐴 a proper subset of 𝐷?
F. Is 𝐷 a proper subset of 𝐴?
G. Is 𝐴 a superset of 𝐷?
H. Is 𝐷 a superset of 𝐴?
I. Is 𝐴 a superset of 𝐶?

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J. Is 𝐶 a superset of 𝐴?
K. Is 𝐴 a proper superset of 𝐶?

A = {x: x is a vowel in the English alphabet} = {A, E, I, O, U} has cardinality 5.

B= {y: y is a letter in the English alphabet} = {A, B, C,. . . , X, Y, Z} has cardinality 26.

Note that every member of set A is also a member of set B.

Hence, A is called a subset of B.
And B is called a superset of A.
Also, note that B has some elements which are not present in A.
Hence, A is called a proper subset of B.
And B is called a proper superset of A.
Definition - Subsets
➢
If a set B has at least one element not in its subset A
➢ A is a proper subset of B.

1-I
Consider the sets {A: 1, 2, 3}, {B: 1, 2, 3, 4} and {C: 1, 2, 3, 4}
Q1: Identify subset and proper subset relationships among the sets above.
S1: A is a subset of B and C
A is also a proper subset of B
B is a subset of C, and C is a subset of B
Q2: State the cardinality of each set
S2: A, B and C have cardinality 3, 4 and 4 respectively.

1.2 Classifying Numbers

A. Natural Numbers / Counting Numbers
Natural numbers are numbers like
ℕ = {1,2,3, … }
Where:
ℕ 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑎𝑚𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
Think of adding 1 to any natural number, howsoever large. Since this is always possible, the number of natural
numbers is infinite.

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B. Whole Numbers
Adding zero to the natural numbers gives us the whole numbers. Numbers
While it may seem very simple:
➢ Zero adds properties that no natural number has.
Specifically, zero satisfies the additive identity: Continuous Discrete
𝑥+𝑎 =𝑥 ⇒𝑎 =0
The kind of equation above is only satisfied by zero.
Real Rational
Hence, whole numbers are numbers like Numbers Numbers
𝕎 = {0,1,2,3, … }
Where:
Integers
𝕎 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑎𝑚𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑊ℎ𝑜𝑙𝑒 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
C. Integers
If you add the negative version of natural numbers to the whole Whole
numbers, you get integers. Numbers
The negative of zero is zero. Hence, there is no need to add a
negative of zero. Natural
Numbers
Hence, positive whole numbers, negated whole numbers and zero
together make up the integers:
𝕀 = {… , −3, −2, −1,0,1,2,3, … }
Where:
𝕀 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑎𝑚𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑊ℎ𝑜𝑙𝑒 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

But there is a small problem. We are going to use the small letter
𝑖
For complex numbers when we reach there. And to avoid potential confusion, mathematicians prefer to use:
ℤ = {… , −3, −2, −1,0,1,2,3, … }

However, 𝕀 is still correct. Just not preferred.

D. Rational Numbers
So far, we have not looked at fractions. None of the numbers that we talked about could handle fractions. Now,
we introduce numbers that are defined as fractions.

A rational number is a number

𝑝
➢ that can be written in the form 𝑞 (or, in other words, as a fraction)
However, there are some important restrictions:
➢ 𝑝 and 𝑞 are integers
➢ 𝑞 ≠ 0, which is very important since we cannot divide by zero, ever.

We can write out this definition in terms of sets, more formally, like this:

1.37: Rational Numbers

The set ℚ represents rational numbers. They are of the form:

𝑝
ℚ = {𝑥 ⏟| 𝑥 = , 𝑝, 𝑞 ∈ ℤ, 𝑞 ≠ 0}
𝑠𝑢𝑐ℎ
𝑞
𝑡ℎ𝑎𝑡

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Example 1.38
Show that √2 is irrational.

Assume to the contrary that √2 is rational.

Then for some 𝑝 and 𝑞:
𝑝
√2 = , 𝑝, 𝑞 ∈ ℤ, 𝑞≠0
𝑞
Square both sides:
𝑝2
2=
𝑞2
In any perfect square, the power of any prime number must be even.

Numerator is a perfect square.

Denominator is also a perfect square.

𝑝2
Therefore, the power of 2 in 𝑞2 must be even.
But, on the LHS, the power of 2 is odd.
Contradiction.
Hence, our original assumption was wrong.

Hence,
√2 𝑖𝑠 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙
E. Irrational Numbers
Numbers that are not rational are irrational.
ℙ = ℚ′ = {𝑥|𝑥 𝑖𝑠 𝑛𝑜𝑡 𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙}

For example, the number 𝜋 which you might have seen in the formula for the area of a circle
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎 𝐶𝑖𝑟𝑐𝑙𝑒 = 𝜋𝑟 2
Is irrational.
Note that
22
𝜋 ≈ 3.14, 𝜋 ≈
7
But 𝜋 is not actually equal to any of these numbers:
22
𝜋 ≠ 3.14, 𝜋 ≠
7
In fact, 𝜋 cannot be represented as a fraction, with integer numerators and denominators.
This is precisely why it is an irrational number.
F. Real Numbers
Real numbers combine rational numbers and irrational numbers.
We use the letter ℝ for the set of real numbers.
ℝ= ℚ
⏟ ∪ ⏟′
ℚ
𝑅𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝐼𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙
𝑁𝑢𝑚𝑏𝑒𝑟𝑠 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

1.39: Real Numbers

Real numbers are any number on the real number line.

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There is no easy way to define real numbers other than this definition.
G. Summary
𝑆𝑒𝑡 𝑜𝑓 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 = ℕ = {1, 2, 3, 4. . . }
𝑆𝑒𝑡 𝑜𝑓 𝑊ℎ𝑜𝑙𝑒 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 = 𝕎 = {0, 1, 2, 3, 4. . . }
𝑆𝑒𝑡 𝑜𝑓 𝐼𝑛𝑡𝑒𝑔𝑒𝑟𝑠 = ℤ = {… , −3 , −2, −1, 0, 1, 2, 3, . . . }
𝑝
𝑅𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 = ℚ = {𝑥: 𝑥 ∈ , 𝑝, 𝑞 ∈ ℤ, 𝑞 ≠ 0}
𝑞
𝐼𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 = ℚ′ = ℙ
𝑅𝑒𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 = ℝ = ℚ
⏟ ∪ ⏟′
ℚ
𝑅𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝐼𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙
𝑁𝑢𝑚𝑏𝑒𝑟𝑠 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

H. Sets and Subsets

N is a subset of W or
ℕ⊂𝕎
W is a subset of Z or
𝕎⊂ℤ
Z is a subset of Q or
ℤ⊂ℚ
Q is a subset of R or
ℚ⊂ℝ
Combine the above:
ℕ⊂𝕎⊂ℤ⊂ℚ⊂ℝ

Q is a superset of Z or 𝑄 ⊃ 𝑍
Z is a superset of W or Z ⊃ 𝑊
W is a superset of N or 𝑊 ⊃ 𝑁
Combine the above:
ℝ⊃ℚ⊃ℤ⊃𝕎⊃ℕ

I. Discrete and Continuous Values

The domain of a function can be of two types: discrete and continuous.

Values that are meaningful only in jumps are called discrete. Some values can take only whole numbers. Some
can take integers. Some can take rational numbers. These are all discrete values.

Values that can be any number on the real number line are called continuous.

Concept 1.40
Classify the following as discrete or continuous. If discrete, then classify as whole number, integer or rational
A. No. of Olympic Contestants in a sport
B. Money expressed in Rupees, to two decimal places

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Part A Numbers
➢ Must be a whole number such as 42 or 5
3
➢ Cannot be fractional or decimal: 5 = 0.6 contestants does not make sense.
Continuous Discrete
➢ Cannot be negative. −3 𝑐𝑜𝑛𝑡𝑒𝑠𝑡𝑎𝑛𝑡𝑠 in a sport does not make sense
Real Rational
Numbers Numbers

Integers

Whole
Part B Numbers
➢ Is a discrete value
➢ Can take 33, or take 13.05
➢ But cannot take 23.001
Numbers
➢ Hence, it is not continuous

Continuous Discrete

1.41: Set Notation for Discrete Values Real Rational

Discrete values are represented using set notation. Numbers Numbers

Concept 1.42 Integers

Write the odd numbers from 13 to 53 in set notation:
Whole
Numbers
Roster Form
We can write all the numbers in a set, as below, and it would be correct:
{13, 15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,51,53}

Roster Form with Ellipsis

But this is a really long way of writing it. So, a shorter way is to
➢ Write enough numbers that the pattern is clear
➢ Put three dots to indicate that the pattern continues
➢ Put the ending number so that we know where to stop
And this can be done as:

{13,15, … ,51,53}

The three dots … are called an ellipsis.

Set Builder Notation

We can also write a rule that the numbers in the set must satisfy, and this is given below:

𝑥 ⏟| 13 ≤ 𝑥 ≤ 53, 𝑥 ⏟
∈ ⏟
ℤ
𝒔𝒖𝒄𝒉 𝒃𝒆𝒍𝒐𝒏𝒈𝒔 𝑺𝒆𝒕 𝒐𝒇
{ 𝒕𝒉𝒂𝒕 𝒕𝒐 𝑰𝒏𝒕𝒆𝒈𝒆𝒓𝒔 }

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J. Rational Values

Concept Example 1.43

A bank calculates whether an account is in overdraft (negative balance) by using the function 𝑓(𝑏), where b is
the balance in dollars. The bank system shows the value dollars and whole cents, which is used as in input for
the function.
The minimum value that a bank account can have is 1 million dollars overdraft, while the maximum value of a
single bank account is five million dollars.
Give the domain of 𝑓(𝑏).

Roster Form with Ellipsis

{−1000000, −999999.99, … ,5000000}

Set Builder Form

𝑝
{𝑥| 𝑥 = , 𝑝, 𝑞 ∈ ℤ, 𝑞 ≠ 0}
𝑞
We can use the definition of rational numbers to write the above set in set builder form, by using a similar style:
𝑝
{𝑥| , −100,000,000 ≤ 𝑝 ≤ 500,000,000, 𝑝 ∈ ℤ}
100

K. Continuous Values
Values like temperature, time, height of a person are continuous values. These values do not have restrictions, in
that they can take all values on the number line.

For example, consider a time period between 10.00 am to 11.00 am. It will take all values on the number line
between 10 am and 11 am.

Concept 1.44
In a cold day in Washington, it is 0° Celsius at 8 am in the morning. By 9 am, the temperate has increased to 10°
Celsius. Was there some point of time between 8 am and 9 am, that the temperature was 𝜋° Celsius?

Yes. There has to be some point of time between 8 am and 9 am that the temperature was 𝜋° Celsius.
The temperature cannot go from 0 to 10, without passing through the value 𝜋.

Concept Example 1.45

𝑃(ℎ) gives the air pressure at a height ℎ feet above sea level in millibars. The height can range from 100 feet
above sea level to 4000 feet above sea level. State the domain of 𝑃(ℎ).

[100,4000]

L. Interval to Inequality Notation (Continuous Values)

Continuous values are represented using sets or intervals on the number line. In general, they are referred to as
interval notation (and not set notation).

If the endpoints of an interval are included, then we use square brackets.

If the endpoints of the interval are not included, then we use round brackets.

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Concept 1.46
Explain the following verbally. Also, write it in interval notation
A. 3 < 𝑥 < 5
B. 3 ≤ 𝑥 ≤ 5
C. 3 ≤ 𝑥 < 5
D. 3 < 𝑥 ≤ 5

3 < 𝑥 < 5 ⇔ (3,5)

3 ≤ 𝑥 ≤ 5 ⇔ [3,5]
3 ≤ 𝑥 < 5 ⇔ [3,5)
3 < 𝑥 ≤ 5 ⇔ (3,5]

M. Choosing between Discrete and Continuous Notations

Example 1.47
Shyam wants to purchase two chocolates for each student in his class for his birthday (including himself). Let 𝑠
be the number of students present in the class of 20 students (including Shyam).

Function Domain
I 𝑓(𝑠) = 2𝑠 𝑠 ∈ (1,20)
II 𝑓(𝑠) = 2𝑠 + 1 𝑠 ∈ [1,20]
III 𝑓(𝑠) = 2𝑠 + 2 𝑠 ∈ {1,2, … ,20}
IV 𝑓(𝑠) = 2𝑠 − 1 𝑠 ∈ {0,1, … ,19}

You need to write all valid pairs comprising a function and a domain from the table above. Use the row number
to identify the functions and the domains.

Answer
The valid pairs are:
𝐼 𝑎𝑛𝑑 𝐼𝐼𝐼
𝐼𝐼𝐼 𝑎𝑛𝑑 𝐼𝑉
1.3 Operations on Sets
A. Intersection and Union

1.48: Intersection of two Sets

The intersection of two sets 𝑋 and 𝑌 is the set of the elements that belong to both 𝑋 and 𝑌.
𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑋 𝑎𝑛𝑑 𝑌 = 𝑋 ∩ 𝑌

1.49: Union of two Sets

The union of two sets 𝑋 and 𝑌 is the set of the elements that belong to either 𝑋 or 𝑌.
𝑈𝑛𝑖𝑜𝑛 𝑜𝑓 𝑋 𝑎𝑛𝑑 𝑌 = 𝑋 ∪ 𝑌

Example 1.50
Let
𝑋 = {𝑥: − 2 < 𝑥 ≤ 4, 𝑥 ∈ ℤ}, 𝑌 = {𝑥: 1 < 𝑥 ≤ 5, 𝑥 ∈ ℤ}
List the elements of:
A. 𝑋

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B. 𝑌
C. 𝑋 ∪ 𝑌
D. 𝑋 ∩ 𝑌

𝑋 = {−1,0,1,2,3,4}
𝑌 = {2,3,4,5}
𝑋 ∪ 𝑌 = {−1,0,1,2,3,4,5}
𝑋 ∩ 𝑌 = {2,3,4}

1.51: Union of Null Set

Union of null set with any set is always the other set
𝜙∪𝑋 =𝑋

1.52: Intersection of Null Set

Intersection of null set with any set is always the null set:
𝜙∩𝑋 =𝜙

1.53: Universal Set

The set of all objects under consideration in a particular scenario is called the universal set. The universal set
can be:
➢ Explicitly defined or given in the question
➢ Understood from the context

1.54: Complement of a Set

All elements in the universal set, but not present in a set are the elements of the complement of the set.
𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑋 = 𝑋 ′
(𝑅𝑒𝑎𝑑: 𝑋 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑂𝑅 𝑋 𝑝𝑟𝑖𝑚𝑒)

Example 1.55
Let the universal set be U = {1,2,3,4,5} and let the sets 𝑋 = {1,2,3} and 𝑌 = {2,4} be defined. Find:
A. 𝑋 ′
B. 𝑌 ′
C. 𝑋 ′ ∩ 𝑌 ′
D. (𝑋 ∩ 𝑌)
E. (𝑋 ∩ 𝑌)′
F. 𝑋 ′ ∪ 𝑌′

𝑋 ′ = {4,5}
𝑌 ′ = {1,3,5}
𝑋 ′ ∩ 𝑌 ′ = {5}
(𝑋 ∩ 𝑌) = {2}
(𝑋 ∩ 𝑌)′ = {1,3,4,5}
𝑋 ′ ∪ 𝑌 ′ = {1,3,4,5}

U = {A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z}
M = {Consonants}
N = {Letters that have at least one line of symmetry}

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List the elements of:

A. M'
B. N'

1.56: Complement of Universal Set

The complement of the universal set is the null set
𝑈′ = 𝜙

1.57: Complement of Null Set

The complement of the null set is the universal set
𝜙′ = 𝑈

1.58: Union of Universal Set

Union of universal set with any set is always the Universal Set:
𝑈∪𝑋 =𝑈

1.59: Intersection of Universal Set

Intersection of universal set with any set is always the other set
𝑈∩𝑋 =𝑋

Challenge 1.60
Let Universal Set, 𝑈, be the set of Numbers from 1 to 25. Let
𝑃 be the set of primes.
𝑂 be the set of odd numbers.
𝐸 be the set of even numbers.
𝐹 be the set of factors of 60.
𝑀 be the set of multiples of 4.

State, in any appropriate form, the sets:

A. 𝑃′ ∩ 𝑂′
B. 𝑃′ ∩ 𝐹 ′
C. 𝑈′
D. 𝑈 ′ ∪ (𝑃′ ∪ 𝑂′ )
E. 𝐸 ′ ∩ 𝑀′
F. 𝑂′ ∩ 𝐸 ′
G. 𝑂′ ∪ 𝐸′

𝑃′ = {1,4,6,8,9,10,12,14,15,16,18,20,21,22,24,25}
𝑂′ = {2,4,6,8,10,12,14,16,18,20,22,24}
′
𝐹 = {7,8,9,11,13,14,16,17,18,19,21,22,23,24,25}
𝐸 ′ = {1,3,5,7,9,11,13,15,17,19,21,23,25}
′
𝑀 = {1,2,3,5,6,7,9,10,11,13,14,15,17,18,19,21,22,23,25}

Part A
𝑃′ ∩ 𝑂′ = {4,6,8,10,12,14,16,18,20,22,24} = 𝑂′ − {2}
Part B

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{8,9,14,16,18,21,22,24,25}
Part C and D
The complement of the universal set is the null set.
𝑃𝑎𝑟𝑡 𝐶: 𝜙
The
𝑃′ ∪ 𝑂′ = {1,2,4,6,8,9,10,12,14,15,16,18,20,21,22,24,25}
𝜙 ∪ (𝑃′ ∪ 𝑂′ ) = 𝑃′ ∪ 𝑂′
Part E
𝐸 ′ ∩ 𝑀′ = {1,3,5,7,9,11,13,15,17,19,21,23,25}

Example 1.61
Intersection of circle and line

Three cases

Example 1.62
Intersection of parabola and line

Three cases

B. AND, OR Notation

𝑼𝒏𝒊𝒐𝒏: 𝐸𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝐴 𝑶𝑹 𝐵 ⇔ 𝐴 ∪ 𝐵
𝑰𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏: 𝐸𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝐴 𝑨𝑵𝑫 𝐵 ⇔ 𝐴 ∩ 𝐵

C. Venn Diagrams

Example 1.63
Represent the sets below on a Venn Diagram.
𝑂 = {𝑥: 𝑥 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 𝑓𝑟𝑜𝑚 1 𝑡𝑜 12}
𝑃 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑝𝑟𝑖𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑓𝑟𝑜𝑚 1 𝑡𝑜 12}
𝑈 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑤ℎ𝑜𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑓𝑟𝑜𝑚 1 𝑡𝑜 12}

𝑂 = {1,3,5,7,9,11}
𝑃 = {2,3,5,7,11}

Example 1.64
P = Primes from that are one more than a multiple of 4 and less than 50.
M = Numbers that are one less than a multiple of 6, and less than 50.

Represent the above sets on a Venn Diagram (without the Universal Set,)

Example 1.65
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S = Two Digit Numbers, the sum of whose digits is a single digit prime number.
P = Two Digit Numbers, the product of whose digits is a single digit prime number
D = Two Digit Numbers, the difference of whose digits is an even prime number

Represent the above sets on a Venn Diagram.

𝑆 = {20,11,30,21,12,50,41,32,23,41,70,61,52,43,34,25,16}
𝑃 = {21,12,31,13,51,15,17,71}
𝐷= {13,20, 24,35,46,57,68,79, 31,42,53,64,75,86,97}

Example 1.66
T = Numbers one more than a multiple of 3.
F = Numbers one more than a multiple of 4.
V = Numbers one more than a multiple of 5.

Universal set = Numbers from 40 to 70.

Example 1.67
𝐴 = 𝐹𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 48
𝐵 = 𝐹𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 60
𝑈𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙 𝑆𝑒𝑡 = 𝑈 = {𝐹𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 48} ∪ {𝐹𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 60} ∪ {𝐹𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 72}

Represent the above sets on a Venn Diagram.

𝐴 = {1,2,3,4,6,8,12,16,24,48}
𝐵 = {1,2,3,4,5,6,10,12,20,30,60}
𝑈 = {1,2,3,4,5,6,8,9,10,12,16,18,20,30,36,48, 60,72}
D. Shading Venn Diagrams

Example 1.68
Consider the Venn Diagram for two sets 𝑋 and 𝑌.
Shade:
A. 𝑋
B. 𝑌
C. 𝑋 ∩ 𝑌
D. 𝑋 ∪ 𝑌
E. 𝑋′
F. 𝑌'

Example 1.69
➢ First Diagram: A
➢ Second Diagram: B
➢ Third Diagram: 𝐴 ∩ 𝐵 ⇔ 𝐴 𝐴𝑁𝐷 𝐵
➢ Third Diagram: 𝐴 ∪ 𝐵 ⇔ 𝐴 𝑂𝑅 𝐵

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Example 1.70
Consider the Venn Diagram for the intersection of three sets X, Y, Z
Shade:
A. 𝑋
B. 𝑌
C. 𝑍
D. 𝑋 ∩ 𝑌
E. 𝑌 ∩ 𝑍
F. 𝑋 ∩ 𝑍
G. 𝑋 ∩ 𝑌 ∩ 𝑍
H. 𝑋′
I. 𝑌′
J. 𝑍′
K. 𝑋 ∩ (𝑌 ∪ 𝑍)
L. 𝑌 ∩ (𝑋 ∪ 𝑍)
M. 𝑍 ∩ (𝑋 ∪ 𝑌)

E. Algebra with Venn Diagrams

1.71: Complement of a Complement is the set itself

(𝐴′ )′ = 𝐴

Example 1.72
(𝐴′ ∪ 𝐵′)′

(𝐴′ ∪ 𝐵′ )′ = (𝐴′ )′ ∩ (𝐵′ )′ = 𝐴 ∩ 𝐵

Example 1.73
(𝐴 ∪ 𝐵)′

(𝐴 ∪ 𝐵)′ = 𝐴′ ∩ 𝐵′

Example 1.74
(𝐴 ∪ 𝐵′)′

(𝐴 ∪ 𝐵′ )′ = 𝐴′ ∩ 𝐵

From left to right

➢ First Diagram: A

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➢ Second Diagram: B’
➢ Third Diagram: 𝐴 ∪ 𝐵′

And now to find the final answer, we can graph the complement of the third diagram:

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2. COUNTING STRATEGIES
2.1 Enumeration
A. Basics
➢ Enumeration requires counting by listing. Some questions in this section are easy, but some are hard.
You should come back to the ones you find hard later.
➢ Note that we will introduce formulas for counting later. But some exam questions require enumeration,
and some require enumerating cases, combined with formulas.

Example 2.1
Enumerate the number of ways in which three people (𝐴, 𝐵, 𝐶) can be seated in a row.

Start with A Start with B Start with C

ABC BAC CAB
ACB BCA CBA

Example 2.2
Enumerate the number of ways in which four people (𝑃, 𝑄, 𝑅, 𝑆) can be seated.

Start with P Start with Q Start with R Start with S

PQ PR PS QP QR QS RP RQ RS SP SQ SR
PQRS PRQS PSQR QPRS QRPS QSPR RPQS RQPS RSPQ SPQR SQPR SRPQ
PQSR PRSQ PSRQ QPSR QRSP QSRP RPSQ RQSP RSQP SPRQ SQRP SRQP

Example 2.3
Carlos is reading a book with 160 pages. How many page numbers in the book contain the digit zero?

The digit zero could be in the Units Place:

10, 20, 30, 40, 50, 60, 70, 80, 90, 100,110, 120, 130, 140, 150, 160 = 17 Numbers
⏟
Zero in the Units place
The digit zero could be in the Ten’s Place:
100,101,102,103,104,105,106,107,108,109
⏟ = 10 Numbers
Zero in the Tens place

But the number 100 gets used in both.

𝑇𝑜𝑡𝑎𝑙 = 17 + 10 − 1 = 26 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

Example 2.4
What is the number of ways in which we can arrange the letters of:
A. JEE
B. IIIT

We can only choose the location of the letter that is different, giving us:
A. JEE, EJE, EEJ
B. TIII, ITII, IITI, IIIT

Example 2.5
A. Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball

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tournament. Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters
be chosen? (AMC 8 2004/4)
B. I need to choose nine friends to invite to my birthday party from my ten best friends. In how many ways
can I do this?
C. A bag contains four pieces of paper, each labeled with one of the digits 1, 2, 3 or 4, with no repeats.
Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number.
What is the probability that the three-digit number is a multiple of 3? (AMC 8 2007/24)

Part A
Choosing three starters is the same as rejecting one person.
Since there are four people in the team, we can do this in four ways.

Part B
10 𝑊𝑎𝑦𝑠
Part C
123 → 6
124 → 7
134 → 8
234 → 9
2 1
𝑃= =
4 2

Example 2.6
Find the number of handshakes if in a room, every person shakes hand with everyone else, and there are:
A. Two People
B. Three People
C. Four People
D. Five People

A shaking hands with B is the same as B shaking hands with A

If there are n people in the room, any one person can shake hands with the remaining (𝑛 – 1) people

Two People: A, B Three People: A, B, C Four People: A, B, C, D Five People: A, B, C, D, E

A: B A: B, C A: B, C, D A: B, C, D, E
B: C B: C, D B: C, D, E
C: D C: D, E
D: E

Example 2.7
How many triangles are created by drawing 𝑛 lines from 𝑛 collinear points to a point 𝑍 not on the line defined
by the 𝑛 collinear points?

Two Points: A, B Three Points: A, B, C Four Points: A, B, C, D Five Points: A, B, C, D, E

AZB AZB, AZC AZB, AZC, AZD AZB, AZC, AZD, AZE
BZC BZC, BZD BZC, BZD, BZE
CZD CZD, CZE
DZE

Example 2.8
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Victor has a quarter, a dime, a nickel and a penny. He wants to buy an item from the shop and pay with exact
change. How many different ways can he pay the shopkeeper?

Suppose Victor pays with exactly one coin. He has four options:
25 cents
10 cents
5 cents
1 cent

Suppose Victor pays with exactly two coins. This means he has to make pairs of the coins. He has six options:
25+10=35 cents
25+5=30 cents
25+1=26 cents
10+5=15 cents
10+1=11 cents
5+1=6 cents

Suppose Victor pays with exactly three coins. Choosing three coins is the same as not choosing one coin. He can
do it in four ways:
25+10+5=40 Cents
25+10+1=36 Cents
25+5+1= 31 Cents
10+5+1=16 Cents

He can pay with all four coins, which can be done in 1 Way:
25+10+5+1=46 Cents

Suppose you want to check whether you have missed any option. Note that there are four coins. Each coin can
be used, or not used. There are two choices for each coin.
Total Choices=2^4=16

Out of which the choice where you use no coins is not applicable, since you are not paying any money to the
shopkeeper. Hence, we have 15 valid choices.
B. Number of Power Sets
A well-defined collection of elements is called a set. A subset of a set A is a set that does not contain any
element not in the set A:
𝐴 = {2, 3, 4} 𝑖𝑠 𝑎 𝑠𝑢𝑏𝑠𝑒𝑡 𝑜𝑓 𝐵 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟}

2.9: Power Set

The power set of A is the set of all subsets of A.
Null set means the set has no elements. The symbol ϕ is used to indicate this.

Example 2.10
What is the number of elements of the power set of the following sets: {𝜙}, {1}, {1, 2}, {1, 2, 3}

No. of Elements {ϕ} {1} {1, 2} {1, 2, 3}

0 {ϕ} {ϕ} {ϕ} {ϕ}
1 {1} {1}, {2} {1}, {2}, {3}
2 {1, 2} {1, 2}, {1, 3}, {2, 3}

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3 {1, 2, 3}

Total 1 2 4 8
20 = 1 21 = 2 22 = 4 23 = 8

C. Casework
Breaking into cases builds on enumeration.
Here, not only must each choice be listed, but also its choices must be counted.

Many aspects need to be kept in mind when deciding the cases (particularly for difficult questions):
➢ Comprehensiveness: Cases cover all possible events
➢ Overcounting: None of the cases, or parts of the cases, cover events twice. If they do, then accounting for
that.
➢ Approach: The “right” approach can make work much easier. This comes only from experience and
practice.
➢ Formula: Choosing the right formula to apply to each case.
➢ Handling edge cases: Cases that need special treatment because of their special situation are handled
appropriately.

In situations where we need to count, we may need to break down the possibilities into cases.
The number of times that something happens is then the sum of all cases.

Example 2.11
My clock chimes two times 15 minutes after the hour, four times 30 minutes after the hour and six times 45
minutes after the hour. The clock also chimes eight times on each hour in addition to chiming the number of
times equal to the hour. (So, at 2:00 p.m., the clock chimes 8 + 2 = 10 times.) Starting at 12:05 a.m., how many
times does the clock chime in a 24-hour period? (Mathcounts 2007 Warm-Up 14)

We use casework
Case I: Chimes that don’t change when the hour changes.

24 ( ⏟
2 + ⏟
4 + ⏟
6 + ⏟
8 ) = 24(20) = 480
𝟏𝟓 𝑴𝒊𝒏𝒖𝒕𝒆𝒔 𝟑𝟎 𝑴𝒊𝒏𝒖𝒕𝒆𝒔 𝟒𝟓 𝑴𝒊𝒏𝒖𝒕𝒆𝒔 𝑶𝒏 𝒕𝒉𝒆 𝑯𝒐𝒖𝒓
Case II: Chimes that change when the hour changes.
12 × 13
2 × (1
⏟ + 2 + 3 + ⋯ + 12) = 2 ( ) = (12 × 13) = 156
𝑻𝒊𝒎𝒆𝒔 𝑬𝒒𝒖𝒂𝒍 𝒕𝒐 𝒕𝒉𝒆 𝑯𝒐𝒖𝒓
2
Add the two cases:
480 + 156 = 636

Example 2.12
The number 64 has the property that it is divisible by its units digit. How many whole numbers between 10 and
50 have this property? (AMC 8 2000)

We break into cases based on the units digit. Each case must be considered separately.
⏟
4 + ⏟
4 +⏟ 1+ ⏟ 2 + ⏟
4 +⏟1+ ⏟ 0 +⏟ 1+ ⏟ 0 = 17
𝟏𝟏,𝟐𝟏,𝟑𝟏,𝟒𝟏 𝟏𝟐,𝟐𝟐,𝟑𝟐,𝟒𝟐 𝟑𝟑 𝟐𝟒,𝟒𝟒 𝟏𝟓,𝟐𝟓,𝟑𝟓,𝟒𝟓 𝟑𝟔 𝟕:𝑵𝒐𝒏𝒆 𝟒𝟖 𝟗:𝑵𝒐𝒏𝒆

Example 2.13
An ascending integer is one in which each digit is greater than any other digit which precedes it (Example: 359).

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How many ascending integers are there between 200 and 300? (NMTC Primary/Final, 2010/8)

All numbers between 200 and 300 will have hundreds digit 2. The lowest value of the ten’s digit will be 3.

Ten’s Digit One’s Digit Number

3 4,5,6,7,8,9 6
4 5,6,7,8,9 5
5 6,7,8,9 4
6 7,8,9 3
7 8,9 2
8 9 1
9 0
Total 21

Example 2.14
For how many three-digit whole numbers does the sum of the digits equal 25? (AMC 8 1994/8)

If the three digits have maximum value, then the sum will be
9 + 9 + 9 = 27
Sum 25 can be accomplished in two ways:
➢ Reduce two from a single digit, giving {7,9,9}, which can be arranged to get
✓ 997, 979, 799
➢ Reduce one from two digits, giving {8,8,9}
✓ 889, 898, 988
In all, we get 6 numbers.

Shortcut
We can also visualize the number of ways to arrange {7,9,9} by saying that we have three places for the 7, and
the remaining numbers give us no choice.
Similarly, there are 3 ways to arrange the 9 in {8,8,9}
𝑇𝑜𝑡𝑎𝑙 𝑊𝑎𝑦𝑠 = 3 + 3 = 6

Example 2.15
How many different four-digit numbers can be formed be rearranging the four digits in 2004? (AMC 8 2004/2)

There are exactly two choices for the first digit:

2 𝑎𝑛𝑑 4
Once we choose the first digit, the second non-zero digit can occupy any of the remaining three places, giving us
numbers:
{2004,2040,2400}, {4002,4020,4200}
This gives us a total of six numbers.

Core Concept 2.16

What is the number of ways of arranging the letters of the word CAT?
{𝐶𝐴𝑇, 𝐶𝑇𝐴, 𝑇𝐴𝐶, 𝑇𝐶𝐴, 𝐴𝑇𝐶, 𝐴𝐶𝑇}

Core Concept 2.17

What is the number of ways of arranging the letters of the word EXAM?

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From the previous question, the number of ways of arranging three letters is 6.
I have 4 choices for the first letter: {𝐸, 𝑋, 𝐴, 𝑀}
With choice of each first letter, I can make six three-letter arrangements of the remaining three letters.

So, total number of choices:

⏟
6 + ⏟
6 + ⏟
6 + ⏟
6 = ⏟
6 × ⏟
4 = 24
𝑭𝒊𝒓𝒔𝒕 𝑭𝒊𝒓𝒔𝒕 𝐹𝑖𝑟𝑠𝑡 𝑭𝒊𝒓𝒔𝒕 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 𝒇𝒐𝒓 𝑨𝒓𝒓𝒂𝒏𝒈𝒎𝒆𝒏𝒕𝒔 𝒇𝒐𝒓
𝑳𝒆𝒕𝒕𝒆𝒓:𝑬 𝑳𝒆𝒕𝒕𝒆𝒓:𝑿 𝐿𝑒𝑡𝑡𝑒𝑟:𝐴 𝑳𝒆𝒕𝒕𝒆𝒓:𝑴 𝑭𝒊𝒓𝒔𝒕 𝑳𝒆𝒕𝒕𝒆𝒓 𝒆𝒂𝒄𝒉 𝒍𝒆𝒕𝒕𝒆𝒓

2.2 Geometrical Counting

A. Counting Squares

Example 2.18
Count the number of squares (of any size) in the diagram alongside.

Squares of size 1. We should be able to get 4 such squares.

𝑇𝑜𝑝 𝐿𝑒𝑓𝑡, 𝑇𝑜𝑝 𝑅𝑖𝑔ℎ𝑡, 𝐵𝑜𝑡𝑡𝑜𝑚 𝐿𝑒𝑓𝑡, 𝐵𝑜𝑡𝑡𝑜𝑚 𝑅𝑖𝑔ℎ𝑡

Squares of size 2. We will get a single such square.

Total number of squares is:

4+1=5

Example 2.19
The diagram alongside shown 9 lattice points. Lattice points means that they are equally
spaced on a coordinate plane.
Count the number of squares that can be formed such that four of the lattice point form
the four vertices of the square.

Squares of
𝑆𝑖𝑧𝑒 1 = 4
𝑆𝑖𝑧𝑒 2 = 1
𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑠 = 1

Total
=4+1+1=6

Example 2.20
Count the number of squares (of any size) in the diagram alongside.

Squares of size 1
9

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Squares of size 2:
4

Squares of size 3:
1

Total
9 + 4 + 1 = 14

Example 2.21
Count the number of squares (of any size) in the diagram alongside.

Squares of size:
𝑆𝑖𝑧𝑒 1: 16
𝑆𝑖𝑧𝑒 2: 9
𝑆𝑖𝑧𝑒 3: 4
𝑆𝑖𝑧𝑒 4: 1

= 16 + 9 + 4 + 1 = 30

Example 2.22
Using the above examples generalize to find the number of squares in a for a 𝑛 × 𝑛 grid of
the type alongside. The diagram has 3 × 3, but we want a general answer.

Squares of size 𝑛

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1
Squares of size 𝑛 − 1
22
Squares of size 𝑛 − 2
32

And so on until squares of size 𝑛

𝑛2

Add it all to get:

𝑛(𝑛 + 1)(2𝑛 + 1)
12 + 22 + ⋯ + 𝑛 2 =
6

B. Counting Rectangles

Example 2.23
Count the number of rectangles in the diagram alongside.

First, count the number of squares:

𝑆𝑖𝑧𝑒 1 × 1 = 4
𝑆𝑖𝑧𝑒 2 × 2 = 1

Then, count the number of rectangles:

𝑆𝑖𝑧𝑒 1 × 2 = 𝑆𝑖𝑧𝑒 2 × 1 = 2

Total
=4+1+2+2=9

Example 2.24
Count the number of rectangles in the diagram alongside.

First, count the number of squares:

𝑆𝑖𝑧𝑒 1 × 1 = 9
𝑆𝑖𝑧𝑒 2 × 2 = 4
𝑆𝑖𝑧𝑒 3 × 3 = 1

Then, count the number of rectangles:

𝑆𝑖𝑧𝑒 1 × 2 = 𝑆𝑖𝑧𝑒 2 × 1 = 6
𝑆𝑖𝑧𝑒 1 × 3 = 𝑆𝑖𝑧𝑒 3 × 1 = 3
𝑆𝑖𝑧𝑒 2 × 3 = 𝑆𝑖𝑧𝑒 3 × 2 = 2

Total
= 9 + 4 + 1 + 6(2) + 3(2) + 2(2) = 36

C. Counting Triangles

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Example 2.25
The diagram alongside consists of a larger triangle split into smaller ones. Count the number
of triangles of any size in the diagram.

Triangles made of
𝑂𝑛𝑒 𝑁𝑢𝑚𝑏𝑒𝑟: 1,2,3,4,5,6 ⇒ 6 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠
𝑇𝑤𝑜 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 12, 14,25,45 ⇒ 4 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒
𝑇ℎ𝑟𝑒𝑒 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 143, 256 ⇒ 2 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒
𝐹𝑜𝑢𝑟 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 123456 ⇒ 1 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒

The total number of triangles is:

6 + 4 + 2 + 1 = 13 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠

D. Geometrical Counting

Example 2.26
The diagram alongside shows a regular pentagon with its diagonals drawn.
Determine the number of triangles in the figure.

Case I: 5 Blue and 5 Green Triangles

The blue triangles have their apex at a vertex of the pentagon, and their base is part
of the diagonals. There are five of them (see diagram).
The green triangles have a side of the pentagon as their base, and the nearest vertex
of the inner pentagon as the apex.

Case II: 5 Brown Triangles

These triangles have a side of the pentagon as a base, and a
vertex of the pentagon as the apex.

Two of them are shown in the diagram alongside. In all, these

are:
Δ𝐸𝐶𝐵, Δ𝐷𝐴𝐵, Δ𝐶𝐸𝐴, Δ𝐵𝐷𝐸, Δ𝐴𝐷𝐶

Case III: 5 Violet Triangles

These triangles have one side of the pentagon as the
base, and one vertex of the inner pentagon 𝐹𝐺𝐽𝐼𝐻 as the
third vertex.
The left diagram shows two of the five triangles, and the
right diagram shows the other three.
In all, they are:
Δ𝐸𝐼𝐴, Δ𝐴𝐽𝐵, Δ𝐼𝐵𝐶, Δ𝐽𝐶𝐷, Δ𝐷𝐺𝐸

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Case IV: 5 Orange Triangles

Like the violet triangles, these triangles have one side of
the pentagon as the base, and one vertex of the inner
pentagon 𝐹𝐺𝐽𝐼𝐻 as the third vertex. They are different
from the violet triangles because they connect to the
vertex on the left, rather than the vertex on their right.
The diagram on the left shows two of the five triangles,
and the one on the right shows the other three.

Case V: 5 Pink Triangles

There are 5 triangles with a diagonal of the pentagon as a base, and a vertex of the pentagon as the apex. These
are shown in the diagram.

Case VI: 5 Grey Triangles

The grey triangles (like the pink triangles) also have a diagonal as the base. However, their apex is on a vertex of
the inner pentagon.

10 × 1 + 5 × 5 = 10 + 25 = 35

Example 2.27
How many squares are there altogether in this diagram? (NMTC Primary/Screening
2005/13)

Case I: Squares of Side Length 1

There are 8 such squares.

Case II: Squares of Side Length 2

The final answer is

8 + 3 = 11 𝑆𝑞𝑢𝑎𝑟𝑒𝑠

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Example 2.28
If we have sticks of the same color and
same length, we can make one triangle
using them. If we have sticks of same length
but two different colors, say blue and red,
we can make four triangles as shown in the
diagram. List out (and count) the triangles can be formed using sticks of same length, but three different colors,
say Red, Blue and Green. (NMTC Primary-Final, Primary-III)

First, note from the diagram:

➢ All colors do not need to be used. The first triangle uses only blue colors.
➢ The third triangle 𝐵𝐵𝑅 can be rotated to give a different triangle, but this is not counted. Hence,
triangles that look the same under rotation are counted as one (and not two triangles).

We split this into cases:

Case I: Triangles using exactly one color:
{𝐵𝐵𝐵, 𝐺𝐺𝐺, 𝑅𝑅𝑅} ⇒ 3 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠

Case II: Triangles using exactly two colors:

One color has to occur twice, and one color has to occur once:
{𝐵𝐵𝐺, 𝐵𝐵𝑅, 𝐺𝐺𝐵, 𝐺𝐺𝑅, 𝑅𝑅𝐵, 𝑅𝑅𝐺} ⇒ 6 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠

Case III: Triangles using exactly three colors:

{𝑅𝐺𝐵} ⇒ 1 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒

3 + 6 + 1 = 10

Example 2.29: Coloring

Each square in a 2 × 2 table is colored either black or white. List out (and count) the different colorings of the
table are there? (NMTC Primary-Final, 2004/11)

Note that the table has four cells. We build up the solution by considering smaller cases first.

Case I:
Suppose there were only one cell:
{𝐵, 𝑊} ⇒ 2 𝑂𝑝𝑡𝑖𝑜𝑛𝑠

Case II:
If there are two cells, the first cell can be either black or white, and the second cell can make use of the options
listed above.

{𝐵𝐵, 𝐵𝑊, 𝑊𝐵, 𝑊𝑊} ⇒ 4 𝑂𝑝𝑡𝑖𝑜𝑛𝑠

Case III:
The first cell can be black or white and the other two cells will have the same choices as Case II

{𝐵𝐵𝐵, 𝐵𝐵𝑊, 𝐵𝑊𝐵, 𝐵𝑊𝑊, 𝑊𝐵𝐵, 𝑊𝐵𝑊, 𝑊𝑊𝐵, 𝑊𝑊𝑊} ⇒ 8 𝑂𝑝𝑡𝑖𝑜𝑛𝑠

Finally, consider there are four cells. The first cell can be black, or can be white:

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{𝐵𝐵𝐵𝐵, 𝐵𝐵𝐵𝑊, 𝐵𝐵𝑊𝐵, 𝐵𝐵𝑊𝑊, 𝐵𝑊𝐵𝐵, 𝐵𝑊𝐵𝑊, 𝐵𝑊𝑊𝐵, 𝐵𝑊𝑊𝑊}

{𝑊𝐵𝐵𝐵, 𝑊𝐵𝐵𝑊, 𝑊𝐵𝑊𝐵, 𝑊𝐵𝑊𝑊, 𝑊𝑊𝐵𝐵, 𝑊𝑊𝐵𝑊, 𝑊𝑊𝑊𝐵, 𝑊𝑊𝑊𝑊}

16 𝑂𝑝𝑡𝑖𝑜𝑛𝑠

2.30: Rectangle
A rectangle is a quadrilateral with all angles 90°.

Example 2.31
How many distinct rectangles can you make with perimeter 16 units and natural number side lengths? There
are two cases here. Rectangles that look the same after rotation can be considered distinct. Or they can be
considered the same. Answer both cases.

Case I: Rotations are distinct

Let
𝑙𝑒𝑛𝑔𝑡ℎ = 𝑎, 𝑤𝑖𝑑𝑡ℎ = 𝑏

𝑃 = 16 ⇒ 2(𝑎 + 𝑏) = 16 ⇒ 𝑎 + 𝑏 = 8
(𝑎, 𝑏) = (1,7), (2,6), (3,5), … , (7,1) ⇒ 7 𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒𝑠

Case II: Rotations are the same rectangle

(1,7), (2,6), (3,5), (4,4), (𝟓, 𝟑), (𝟔, 𝟐)(𝟕, 𝟏) ⇒ 7 − 3 = 4

Example 2.32
How many distinct rectangles can you make with perimeter 17 units and natural number side lengths?
(Information regarding rotation is 𝑚𝑖𝑠𝑠𝑖𝑛𝑔……Why so?)

17
𝑃 = 17 ⇒ 2(𝑎 + 𝑏) = 17 ⇒ 𝑎 + 𝑏 =
2
17
𝑅𝐻𝑆 = 2
is a fraction.
This means the 𝐿𝐻𝑆 must also be a fraction.

If I add two natural numbers, the result is always a natural number.

Hence, there are no solutions to the above.
(𝑎, 𝑏) = 𝜙 ⇒ 𝑁𝑜 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠

Example 2.33
Use the answer to the previous example to count the number of rectangles with integral sides for a perimeter of

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n units, where n is a real number.

Answer for both cases with respect to rotation.

Case I: 𝑛 is not an even number

𝑁𝑜𝑡 𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒
Case II: 𝑛 is even
𝑛
𝑃 = 𝑛 ⇒ 2(𝑎 + 𝑏) = 𝑛 ⇒ 𝑎 + 𝑏 =
2
𝑛 𝑛 𝑛 𝑛
(1, − 1) (2, − 2) , … ( − 1,1) ⇒ − 1 𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒𝑠
2 2 2 2
E. Rectangles: Restrictions

Example 2.34
How many rectangles can you make with perimeter 20 units and side lengths odd integers?

Example 2.35
How many rectangles can you make with perimeter 24 units and side lengths even integers?

F. Rectangles: Inequalities

Example 2.36
How many rectangles can you make with perimeter 16 units or less and integral side lengths.

𝑃 = 4 ⇒ (1,1) ⇒ 1 𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒
𝑃 = 6 ⇒ (1,2), (2,1) ⇒ 2 𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒
𝑃 = 8 ⇒ (1,3), (2,2)(3,1) ⇒ 3 𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒
.
.
.
𝑃 = 16 ⇒ (1,7), (2,6), … , (7,1) ⇒ 7 𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒𝑠

7×8
𝑇𝑜𝑡𝑎𝑙 = 1 + 2 + ⋯ + 7 = = 28
2

𝑎+𝑏 <8
𝑎′ + 𝑏 ′ < 7
𝑎′ + 𝑏 ′ ≤ 6
𝑎 + 𝑏′ + 𝑐 ′ = 6
′

8 7×8
( )= = 28
2 2
G. Counting Triangles

Example 2.37
One angle of a triangle is 60°. The other two angles have integer measures. Find the number of such triangles if
the angles cannot all have the same measure.

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We know that one angle of the triangle is 60°. Hence, the other two angles must add up to:
180 − 60 = 120°
If the angles had been equal, they would have been:
60 + 60 = 120
But the angles cannot be equal. Hence, the above is not a possibility.

Now, find the possibilities. An angle in a triangle cannot be 0°. Hence, the smallest value that an angle can take
is:
1° ⇒ 𝑂𝑡ℎ𝑒𝑟 𝐴𝑛𝑔𝑙𝑒 = 119°

We can tabulate all the possibilities like this:

First Angle 1° 2° 3° . . . 59°
Second Angle 119° 118° 117° . . . 61°

And hence, the number of triangles is:

59

Example 2.38
One angle of a triangle is 60°. The other two angles have integer measures. Find the number of such triangles if
the triangle is acute.

We cannot have a right or an obtuse angle in the triangle. Hence, the possibilities are:

First Angle 31° 32° 33° . . . 60°

Second Angle 89° 88° 87° . . . 60°

And hence, the number of triangles is

60 − 31 + 1 = 30

Example 2.39
One angle of a triangle is 60°. The other two angles have integer measures. Find the number of such triangles if
the other two angles are both even.

First Angle 2° 4° . . . 60°

Second Angle 118° 116° . . . 60°

And hence we have to count the number of numbers in the list:

2,4,6, … ,60
Divide each number by 2 to get:
1,2,3, . . ,30 ⇒ 30 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 ⇒ 𝟑𝟎 𝑻𝒓𝒊𝒂𝒏𝒈𝒍𝒆𝒔

Example 2.40
One angle of a triangle is 60°. The other two angles have integer measures. Find the number of such triangles if
the other angles are both odd.

Use complementary counting

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60
⏟ − 30
⏟ = 𝟑𝟎 𝑻𝒓𝒊𝒂𝒏𝒈𝒍𝒆𝒔 𝒘𝒊𝒕𝒉 𝑻𝒘𝒐 𝑶𝒅𝒅 𝑨𝒏𝒈𝒍𝒆𝒔
𝑇𝑜𝑡𝑎𝑙 𝐸𝑣𝑒𝑛
𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠 𝐴𝑛𝑔𝑙𝑒𝑠

First Angle 1° 3° . . . 59°

Second Angle 119° 117° . . . 61°

And hence we have to count the number of numbers in the list:

1,3,5, … ,59
Add 1 to each number to get:
2,4,6, … ,60
Divide each number by 2 to get:
1,2,3, . . ,30 ⇒ 30 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 ⇒ 𝟑𝟎 𝑻𝒓𝒊𝒂𝒏𝒈𝒍𝒆𝒔

Example 2.41
One angle of a triangle is 60°. The other two angles have integer measures. Find the number of such triangles if
the other two angles are both even, and all angles are acute.

First Angle 32° 34° . . . 60°

Second Angle 88° 86° . . . 60°

And hence we have to count the number of numbers in the list:

32,34, … ,60
Divide each number by 2 to get:
16,17, . . ,30 ⇒ 30 − 16 + 1 = 15 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 ⇒ 𝟏𝟓 𝑻𝒓𝒊𝒂𝒏𝒈𝒍𝒆𝒔
H. Semiperimeter

2.42: Longest side less then semiperimeter

In a triangle if a is the longest side (and the other sides are b and c):
𝑎+𝑏+𝑐
𝑎 < 𝑠, 𝑠=
2
The longest side in a triangle is less than the semiperimeter.

𝑎+𝑏+𝑐
𝑎<
2

Multiply both sides of the given inequality by 2:

2𝑎 < 𝑎 + 𝑏 + 𝑐
Subtract a from both sides:
𝑎 <𝑏+𝑐
And the above is the triangle inequality (which is true), and the steps are reversible.
Hence, the original inequality is also true.

Example 2.43
You want to make triangles with 𝑛 matchsticks. Matchsticks are all the same. You cannot put a matchstick on top
of another. How many different triangles can you make?
Write the answer for 𝑛 = 1,2,3, …

1 𝑜𝑟 2 𝑜𝑟 4 𝑀𝑎𝑡𝑐ℎ𝑠𝑡𝑖𝑐𝑘𝑠: 𝑍𝑒𝑟𝑜 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠

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3 𝑀𝑎𝑡𝑐ℎ𝑠𝑡𝑖𝑐𝑘𝑠: 1 − 1 − 1 ⇒ 1 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒
5 𝑀𝑎𝑡𝑐ℎ𝑠𝑡𝑖𝑐𝑘𝑠: 2 − 2 − 1 ⇒ 1 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒
6 𝑀𝑎𝑡𝑐ℎ𝑠𝑡𝑖𝑐𝑘𝑠: 2 − 2 − 2 ⇒ 1 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒
7 𝑀𝑎𝑡𝑐ℎ𝑠𝑡𝑖𝑐𝑘𝑠: 3 − 2 − 2 ⇒ 1 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒

2.44: Triangle Inequality

𝑎 <𝑏+𝑐

Example 2.45
How many different isosceles triangles have integer side lengths and perimeter 23? (2005 AMC8)

(1,1,21) ⇒ 1 + 1 = 2 < 21 ⇒ 𝑁𝑜𝑡 𝑉𝑎𝑙𝑖𝑑

.
.
.
(5,5,13) ⇒ 5 + 5 = 10 < 13 ⇒ 𝑁𝑜𝑡 𝑉𝑎𝑙𝑖𝑑
(6,6,11) ⇒ 6 + 6 = 12 > 11, 11 − 6 = 5 < 6 ⇒ 𝑉𝑎𝑙𝑖𝑑
(7,7,11) ⇒ 6 + 6 = 12 > 11, 11 − 6 = 5 < 6 ⇒ 𝑉𝑎𝑙𝑖𝑑
(8,8,11) ⇒ 6 + 6 = 12 > 11, 11 − 6 = 5 < 6 ⇒ 𝑉𝑎𝑙𝑖𝑑
(9,9,11) ⇒ 6 + 6 = 12 > 11, 11 − 6 = 5 < 6 ⇒ 𝑉𝑎𝑙𝑖𝑑
(10,10,11) ⇒ 6 + 6 = 12 > 11, 11 − 6 = 5 < 6 ⇒ 𝑉𝑎𝑙𝑖𝑑
(11,11,1) ⇒ 6 + 6 = 12 > 11, 11 − 6 = 5 < 6 ⇒ 𝑉𝑎𝑙𝑖𝑑

2.3 Complementary Counting

A. Complementary Counting
In complementary counting, we count the items that we don’t and subtract those from the total.

Example 2.46
What is the number of composite numbers less than 100?

There are 25 prime numbers less than 100. Also, the number 1 is neither prime nor composite.
99 − 25 ⏟ − ⏟
1
𝑷𝒓𝒊𝒎𝒆 𝑵𝒆𝒊𝒕𝒉𝒆𝒓 𝑷𝒓𝒊𝒎𝒆
𝑵𝒖𝒎𝒃𝒆𝒓𝒔 𝒂𝒏𝒅 𝑪𝒐𝒎𝒑𝒐𝒔𝒊𝒕𝒆

Example 2.47
How many numbers less than 10000 do not have three factors?

𝑝2 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 {1, 𝑝, 𝑝2 } ⇒ 9999 −

⏟ 25
⏟ = 9975
𝑭𝒐𝒓 𝒑𝒓𝒊𝒎𝒆 𝒑 𝑵𝒖𝒎𝒃𝒆𝒓𝒔 𝒘𝒊𝒕𝒉
𝒕𝒉𝒓𝒆𝒆 𝒇𝒂𝒄𝒕𝒐𝒓𝒔

Example 2.48
How many positive integers less than or equal to 100 have a prime factor that is greater than 4? (MathCounts
2004 Workout 9)

A number that only has prime factors greater than 4 must not have prime factors of 2 or 3. We classify on the
basis of powers of 3:

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Powers Numbers
of 3
30 20 , 21 , … , 26 7
31 3 × 2 , 3 × 21 , … ,3 × 25
0 6
32 32 × 20 , 32 × 21 , … , 32 × 23 4
33 32 × 20 , 32 × 21 2
34 34 1

Total 20

The numbers we want are:

100 − 20 = 80

Instead of listing out in exponent form, we can also list out in number form:
No.
1 2 4 8 16 32 64 7
3 6 12 24 48 96 6
9 18 36 72 4
27 54 2
81 1

2.4 Reframing the Question / Clever Logic

A. Reframing the Question

There are a number of ways to approach a question. The manner in which you approach the question can make
it easier or more difficult. In the questions below, some strategies work for all questions, while some are easy to
apply for specific types of questions only. Choosing the right strategy is very important in arriving at the answer
quickly and correctly.
Rearranging
A standard, powerful, mathematical technique used to prove many results (including important ones) is to:
➢ Show that a complicated expression is equivalent to another, simpler expression
➢ Calculate the simpler expression

Example 2.49
Town A and B are 60 km apart, and are connected by a straight-line track. Train X starts from town A and goes
towards town B. At the same time, train Y starts from town B, and goes toward town A. Also, at the same time, a
crow sitting at the top of train X starts flying towards train Y. As soon as it touches train Y, it turns back and flies
towards train X. It continues this process till the trains meet. (Make simplifying assumptions: the distance
between the two tracks is negligible, the turning time required for the bird is negligible, etc). Assume that each
𝑘𝑚 𝑘𝑚
train has a speed of 60 ℎ𝑟 , and the bird has a speed of 150 ℎ𝑟 . Find the total distance travelled by the bird.

Long Method
𝑘𝑚
𝑆 = 90 , 𝐷 = 60 𝑘𝑚 ⇒ 𝑇𝑟𝑎𝑖𝑛 = 40 𝑘𝑚, 𝐵𝑖𝑟𝑑 = 20 𝑘𝑚
ℎ𝑟

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𝑘𝑚 40 20
𝑆 = 90 , 𝐷 = 20 𝑘𝑚 ⇒ 𝑇𝑟𝑎𝑖𝑛 = 𝑘𝑚, 𝐵𝑖𝑟𝑑 = 𝑘𝑚
ℎ𝑟 3 3

20 20 20
20 + + 2 + 3 +⋯
3 3 3
Short Method
𝐷 60 60 1
𝑇𝑖𝑚𝑒𝐵𝑖𝑟𝑑 = 𝑇𝑖𝑚𝑒𝑇𝑟𝑎𝑖𝑛 = = = = ℎ𝑟 ⇒ 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝐵𝑖𝑟𝑑 = 15 𝑘𝑚
𝑆 60 + 60 120 2
Pure Logic Method
Trains meet in the middle since they have the same speed. Hence, each train covers 30 km.

Speed of Bird is half the speed of the train, but it travels for same time as the trains.
30
∴ 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝐵𝑖𝑟𝑑 = = 15 𝑘𝑚
2

A. Single Elimination Tournaments

Unless otherwise specified:
➢ Matches are played between two competitors or teams
Single Elimination Format
Matches are played between two competitors. Winners move on to play winners of other matches.
Round Robin Format
All competitors play all other competitors once.

Example 2.50
Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament.
The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games
will be played to determine the winner? (AMC 8 2004)

Enumeration:
⏟
8 + ⏟
4 + ⏟
2 + ⏟
1 = 15
𝑷𝒓𝒆−𝑸𝒖𝒂𝒓𝒕𝒆𝒓−𝑭𝒊𝒏𝒂𝒍 𝑸𝒖𝒂𝒓𝒕𝒆𝒓−𝑭𝒊𝒏𝒂𝒍 𝑺𝒆𝒎𝒊−𝑭𝒊𝒏𝒂𝒍 𝑭𝒊𝒏𝒂𝒍
Logic
Every team which leaves the tournament must lose a game. 15 teams will lose, and leave the game. One team
will have a string of wins, and become the winner.
𝑇𝑜𝑡𝑎𝑙 𝐺𝑎𝑚𝑒𝑠 = 𝑇𝑜𝑡𝑎𝑙 𝑁𝑜. 𝑜𝑓 𝐿𝑜𝑠𝑒𝑟𝑠 = 15

Example 2.51
Repeat the previous example, except that instead of 16 teams, suppose that there are 12 teams.
Enumeration:
The enumeration method requires a little more work:
⏟
6 + ⏟
3
𝑷𝒓𝒆−𝑸𝒖𝒂𝒓𝒕𝒆𝒓−𝑭𝒊𝒏𝒂𝒍 𝑸𝒖𝒂𝒓𝒕𝒆𝒓−𝑭𝒊𝒏𝒂𝒍
Now we have three teams, which is an odd number of teams, so one team will have to sit out, and get a bye. We
will play:
⏟
1
𝑺𝒆𝒎𝒊−𝑭𝒊𝒏𝒂𝒍
This will give one winner from the 3rd round, and one team which got a bye, making a total of two teams. These
two teams will play one game to decide the winner:

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⏟
1
𝑭𝒊𝒏𝒂𝒍
𝑇𝑜𝑡𝑎𝑙 = 6 + 3 + 1 + 1 = 11
Logic
The logic doesn’t change and is easy to apply:
𝑇𝑜𝑡𝑎𝑙 𝐺𝑎𝑚𝑒𝑠 = 𝑇𝑜𝑡𝑎𝑙 𝑁𝑜. 𝑜𝑓 𝐿𝑜𝑠𝑒𝑟𝑠 = 12 − 1 = 11

Example 2.52
Two tournaments A and B are played in a single elimination format, with no draws. How many matches will
there be if there are:
A. 64 competitors?
B. 37 competitors?
63, 36

Example 2.53
116 people participated in a singles tennis tournament of knock out format. The players are paired up in the
first round, the winners of the first round are paired up in second round, and so on till the final is played
between two players. If after any round, there is odd number of players, one player is given a bye, i.e. he skips
that round and plays the next round with the winners. Find the total number of matches played in the
tournament. (CAT 1990/72)

116 − 1 = 115

Example 2.54
CAT 2008 – Data Sufficiency

B. Round Robin Tournaments

All competitors play all other competitors once.

C. Self-Referential Counting

Example 2.55
There are ______ vowels in this short sentence.
A. Twelve
B. Thirteen
C. Fourteen
D. Fifteen
E. Sixteen

𝑇𝑤𝑒𝑙𝑣𝑒: 12 + 2 = 14
𝑇ℎ𝑖𝑟𝑡𝑒𝑒𝑛: 12 + 3 = 15
𝐹𝑜𝑢𝑟𝑡𝑒𝑒𝑛: 12 + 4 = 16
𝐹𝑖𝑓𝑡𝑒𝑒𝑛: 12 + 3 = 15
𝑆𝑖𝑥𝑡𝑒𝑒𝑛: 12 + 3 = 15

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3. COUNTING WITH SETS

3.1 Single Set: Lists
A. Introduction

3.1: Lists
A single set has 𝑛 elements. Sets are not ordered. But, we can order the elements of a set using a logical process
to count the elements more easily.

3.2: Position
Any list will have a position for each element. We are often interested in the position of an element, either by
itself, or, more often, as part of a larger problem.

3.3: Inclusive and Exclusive Counting

𝑁𝑢𝑚𝑏𝑒𝑟𝑠 𝐹𝑟𝑜𝑚 3 𝑡𝑜 97 ∈ ℕ ⇔ ⏟
⏟ 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 𝐵𝑒𝑡𝑤𝑒𝑒𝑛 2 𝑎𝑛𝑑 98 ∈ ℕ ⇔ 3, 4, 5, … 96,97
𝐼𝑛𝑐𝑙𝑢𝑠𝑖𝑣𝑒: 3 𝑎𝑛𝑑 97 𝑎𝑟𝑒 𝑖𝑛𝑐𝑙𝑢𝑑𝑒𝑑 𝐸𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒: 2 𝑎𝑛𝑑 98 𝑎𝑟𝑒 𝑒𝑥𝑐𝑙𝑢𝑑𝑒𝑑

➢ Another way to specify whether the endpoints are included or excluded is by explicitly mentioning in
the question whether the counting is inclusive or exclusive.

Anand counts the numbers 10,11, … ,99,100 where 10 is not included, but 100 is included.

B. Terminology
It’s important to pay attention to the kind of numbers which are we being asked to count
➢ Even vs Odd
➢ Integers versus Natural Numbers
➢ ⏟𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝐼𝑛𝑡𝑒𝑔𝑒𝑟𝑠 versus ⏟ 𝑁𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝐼𝑛𝑡𝑒𝑔𝑒𝑟𝑠 versus ⏟
𝑁𝑜𝑛 − 𝑁𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝐼𝑛𝑡𝑒𝑔𝑒𝑟𝑠
{𝟏,𝟐,𝟑… } {…,−𝟑,−𝟐,−𝟏} {𝟎,𝟏,𝟐,𝟑….}
✓ Non-negative includes zero, but positive does not include zero

Example 3.4
What is the number of positive integers between −4 and 14?

𝑛{−4, −3, … 0,1,2 … ,14} = 𝑛{1,2 … 13} = 13

Example 3.5
What is the number of negative integers from −32 to 12?

𝑛{−32, −31, … 𝟎, 𝟏, 𝟐, … 𝟏𝟏, 𝟏𝟐} = 𝑛{−32, −31, … , −2, −1} = 32

Example 3.6
What is the number of non-negative integers from -32 to 12?

𝑛{−𝟑𝟐, −𝟑𝟏, … − 𝟏, 0,1,2, … 11,12} = 𝑛{0,1,2, … ,12} = 13

Example 3.7
What is the positive difference in the number of elements of 𝑋 and Y if:

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A. 𝑋 is the set consisting of all positive integers less than or equal to 𝑛, where 𝑛 is some arbitrary but fixed
natural number.
B. 𝑌 is the set consisting of all nonnegative integers less than or equal to 𝑛, where 𝑛 is some arbitrary but
fixed natural number

𝑋= {1,2,3 … . , 𝑛}
⏟ ,𝑌 = {0,1,2,3, … 𝑛}
⏟ ⇒ (𝑛 + 1) − 𝑛 = 1
𝑭𝒊𝒓𝒔𝒕 𝒏 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒊𝒏𝒕𝒆𝒈𝒆𝒓𝒔 𝑭𝒊𝒓𝒔𝒕 𝒏+𝟏 𝒏𝒐𝒏𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆 𝑰𝒏𝒕𝒆𝒈𝒆𝒓𝒔

C. Number of elements in a list

To count the numbers from 12 to 74, we can add a few numbers and then remove the ones we don’t want:
{12, 13, 14, … ,73,74} ⇔ 𝑛 {𝟏,
⏟ 𝟐, 𝟑, … 𝟏𝟎, 𝟏𝟏 , ⏟
12, 13, 14, … ,73,74} ⇔ 74 − 11 = 63
𝑵𝒐𝒕 𝑪𝒐𝒖𝒏𝒕𝒆𝒅 𝑪𝒐𝒖𝒏𝒕𝒆𝒅
Since we know the first and the last element in the list when we start with the question, it becomes easier to
write the formula in terms of those numbers:
74
⏟ − 12
⏟ + ⏟
1 = 63
𝑬𝒏𝒅 𝑵𝒖𝒎𝒃𝒆𝒓 𝑺𝒕𝒂𝒓𝒕 𝑵𝒖𝒎𝒃𝒆𝒓 𝑨𝒅𝒅 𝒐𝒏𝒆 𝒎𝒐𝒓𝒆

You want to count the numbers from 12 to 74.

➢ Imagine you have people standing in a row, with one person, each holding the numbers from 12 to 74.
➢ Suppose, you replace the number
✓ 12 with 12 − 11 = 1
✓ 13 with 13 − 11 = 2
✓ The number 14 − 11 = 3
✓ .
✓ .
✓ .
✓ The number 74 with 74 − 11 = 63

Start Number End Number

12 13 14 … 74 𝒙 𝒏𝒖𝒎𝒃𝒆𝒓𝒔
Subtract 11 from 11 11 11 … 11
each number
1 2 3 … 74 − 11 𝒙 𝒏𝒖𝒎𝒃𝒆𝒓𝒔

3.8: Bijection Principle

Subtracting 11 from the list above changes the values of the numbers, but not how many numbers it contains.
This same idea is used in the proof below.

➢ A bijection is a one to one mapping between the elements of one set and the elements of another set.
➢ Each element of the first set is mapped to exactly one element of the second set.
➢ Hence, the number of elements in the two sets is exactly the same. In more technical terms, the two sets
have the same cardinality.

3.9: Number of elements in a list starting from 𝒂 and ending at 𝒃

The number of numbers in a list of integers starting from 𝑎, and ending at 𝑏 is
⏟
𝑏 − ⏟
𝑎 +1
𝑬𝒏𝒅 𝑵𝒖𝒎𝒃𝒆𝒓 𝑺𝒕𝒂𝒓𝒕 𝑵𝒖𝒎𝒃𝒆𝒓

You have a list

{𝑎, 𝑎 + 1, 𝑎 + 2, … , 𝑏}

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Where 𝑎 is an integer.
Find the number of numbers in the list.

Start End Number

Number
Subtract 𝒂 − 𝟏 𝑎 𝑎 + 1 𝑎 + 2 ... 𝑏
𝒇𝒓𝒐𝒎 𝒆𝒂𝒄𝒉 𝒏𝒖𝒎𝒃𝒆𝒓 𝒊𝒏 𝒕𝒉𝒆 𝒍𝒊𝒔𝒕 𝑎−1 𝑎−1 𝑎−1 . .. 𝑎−1
1 2 3 . .. 𝑏−𝑎+1
⏟
𝑻𝒐𝒕𝒂𝒍 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑵𝒖𝒎𝒃𝒆𝒓𝒔
𝒊𝒏 𝒕𝒉𝒆 𝒍𝒊𝒔𝒕

𝑎 − (𝑎 − 1) = 𝑎 − 𝑎 + 1 = 1
𝑎 + 1 − (𝑎 − 1) = 𝑎 + 1 − 𝑎 + 1 = 2

𝑏 − (𝑎 − 1) = 𝑏 − 𝑎 + 1

Example 3.10
What is the number of positive integers from 7 to 82?

We want to find the number of numbers in the list:

𝑛{7, 8, … 82}
We add a few numbers to make it complete, and then see which ones we don’t want.
𝑛{1,2,3, … ,6,7,8,9, … ,81,82} − ⏟
⏟ 𝑛{1,2, … 7}
82 7
This process also eliminates one number that we do want (which is 7), so we add it back:
=⏟𝑛{1,2,3, … ,6,7,8,9, … ,81,82} − ⏟
𝑛{1,2, … 7} + 1 = 82 − 7 + 1 = 76
82 7
Shortcut
82
⏟ − ⏟
7 + 1 = 76
𝑬𝒏𝒅 𝑵𝒖𝒎𝒃𝒆𝒓 𝑺𝒕𝒂𝒓𝒕 𝑵𝒖𝒎𝒃𝒆𝒓

Example 3.11
Find the number of
A. Positive Integers from 8 to 54
B. Positive Integers between 8 and 54

Part A
Apply the formula
54
⏟ − ⏟
8 + 1 = 47
𝑬𝒏𝒅 𝑺𝒕𝒂𝒓𝒕
𝑵𝒖𝒎𝒃𝒆𝒓 𝑵𝒖𝒎𝒃𝒆𝒓

Part B
Using the previous answer
The number of numbers which is between 8 and 54 should be two less than the numbers from 8 to 54, and
hence it should be:
47 − 2 = 45
Direct Calculation
For direct calculation, it is best to convert the 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 into 𝑓𝑟𝑜𝑚 − 𝑡𝑜:
𝐵𝑒𝑡𝑤𝑒𝑒𝑒𝑛 8 𝑎𝑛𝑑 54 ⇔ 𝐹𝑟𝑜𝑚 9 𝑡𝑜 53
And then we can use the formula:
53 − 9 + 1 = 45

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Example 3.12
Find the number of
A. Negative Integers from −72 to −33
B. Negative Integers between −72 and −33

Part A
−33 − (−72) + 1 = −33 + 72 + 1 = −33 + 73 = 40
Part B
Convert the 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 into 𝑓𝑟𝑜𝑚 − 𝑡𝑜:
𝐵𝑒𝑡𝑤𝑒𝑒𝑒𝑛 − 72 𝑎𝑛𝑑 − 33 ⇔ 𝐹𝑟𝑜𝑚 − 71 𝑡𝑜 − 34
And then we can use the formula:
−34 − (−71) + 1 = −34 + 71 + 1 = −34 + 72 = 38

Example 3.13
Find the number of positive integers between −72 and −33

The range of integers which has been given is all negative. Hence, the number of positive integers in this range
is
𝑍𝑒𝑟𝑜

Example 3.14
Tom counted from the largest two-digit number to the smallest four-digit number (including both). How many
numbers did he count in all?

1000 − 99 + 1 = 902

Review 3.15: Counting Negative Numbers, From and Between

Tweedledum and Tweedledee were counting integers. Tweedledum counted 𝑋 numbers from −12 to 48.
Tweedledee counted 𝑌 numbers between −13 and 49. They did not skip any number while counting. What is
𝑋𝑌 (where XY refers to the product of 𝑋 and 𝑌)?

Formula
𝑋= 48
⏟ − (−12)
⏟ + 1 = 61 ⇒ 𝑋𝑌 = 612 = 3721
𝑺𝒕𝒂𝒓𝒕 𝑵𝒖𝒎𝒃𝒆𝒓 𝑬𝒏𝒅 𝑵𝒖𝒎𝒃𝒆𝒓

Logic
We want the numbers
{−12, −11, −10, … , −1,0,1, . .47,48}
12
⏟ + ⏟1 + 48
⏟ = 61
𝑵𝒆𝒈𝒂𝒕𝒊𝒗𝒆 𝑵𝒖𝒎𝒃𝒆𝒓𝒔 𝒁𝒆𝒓𝒐 𝑷𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝑵𝒖𝒎𝒃𝒆𝒓𝒔

Tweedledee
For him, we want the numbers
𝐵𝑒𝑡𝑤𝑒𝑒𝑛 − 13 𝑎𝑛𝑑 49
Which is the same as the numbers
𝑓𝑟𝑜𝑚 − 12 𝑡𝑜 48
And hence

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𝑌 = 𝑋 = 61

D. Real Life Scenarios

Example 3.16: Packing Strawberies

Randhir is packing strawberries in a box, with seven strawberries to each box. He starts with the box numbered
seven. He continues packing boxes numbered eight, nine, and so on. The last box he packs is the box numbered
twenty eight. What is the number of strawberries that Randhir that he packs?

⏟
7 × (28
⏟ − 7 + 1) = 7 × 22 = 154
𝑺𝒕𝒓𝒂𝒘𝒃𝒆𝒓𝒓𝒊𝒆𝒔 𝑵𝒐.𝒐𝒇 𝑩𝒐𝒙𝒆𝒔
𝒑𝒆𝒓 𝑩𝒐𝒙

Example 3.17: Counting Seats in Rows

If rows 9 through 21 of a theatre (with each row having 12 seats) are filled, then how many people are seated in
those rows? (Note: Rows 9 through 21 means rows 𝑓𝑟𝑜𝑚 9 𝑡𝑜 21 – 𝑖𝑛𝑐𝑙𝑢𝑑𝑖𝑛𝑔 9 and 21).

(21 − 9 + 1) ×
⏟ 12
⏟ = 13 × 12 = 156
𝑵𝒐.𝒐𝒇 𝑹𝒐𝒘𝒔 𝑺𝒆𝒂𝒕𝒔 𝒑𝒆𝒓 𝑹𝒐𝒘

E. Counting Number of Numbers

Example 3.18: Counting Numbers

What is the number of:
A. Two-digit numbers
B. Three-digit numbers
C. 𝑛-digit Numbers

Parts A and B
{10,11,12, … ,99} = {𝟏,
⏟ 𝟐, 𝟑, … , 𝟗 , ⏟
10,11,12, … . ,99} = 99 − 10 + 1 = 90
𝑫𝒐𝒏′ 𝒕 𝑪𝒐𝒖𝒏𝒕 𝐶𝑜𝑢𝑛𝑡

{100,101,102, … . ,999} = {𝟏,

⏟ 𝟐, 𝟑, … , 𝟗𝟗 , ⏟
100,101,102, … . ,999} = 999 − 100 + 1 = 900
𝑫𝒐𝒏′ 𝒕 𝑪𝒐𝒖𝒏𝒕 𝐶𝑜𝑢𝑛𝑡

Part C

2 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 90 = 9 × 101 = 9 × 102−1

3 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 900 = 9 × 102 = 9 × 103−1
4 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 9000 = 9 × 103 = 9 × 104−1
𝑛 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 9 × 10𝑛−1

3.19: Number of 𝒏 digit Numbers

The number of n digit numbers is
9 × 10𝑛−1

Example 3.20: Counting Numbers

What is the difference between the number of six-digit natural numbers and the number of two-digit natural

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numbers?

9,00,000
⏟ – 90
⏟ = 899,910
𝐍𝐨.𝐨𝐟 𝟔−𝐝𝐢𝐠𝐢𝐭 𝐍𝐨.𝐨𝐟 𝟐−𝐝𝐢𝐠𝐢𝐭
𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐧𝐮𝐦𝐛𝐞𝐫𝐬

F. Counting Possible Totals

Example 3.21: Sum of Dice Rolls

Shane rolls two six-sided dice, with faces numbered from one to six, and adds the numbers that come up on the
two dice. How many such distinct totals can he get?

{1,2,3,4,5,6}
⏟ , {1,2,3,4,5,6}
⏟ ⇒ 𝑀𝑖𝑛 𝑆𝑢𝑚 = 1 + 1 = 2, 𝑀𝑎𝑥 𝑆𝑢𝑚 = 6 + 6 = 12
𝑭𝒊𝒓𝒔𝒕 𝑫𝒊𝒆 𝑺𝒆𝒄𝒐𝒏𝒅 𝑫𝒊𝒆
𝑊𝑒 𝑐𝑎𝑛 𝒂𝒄𝒉𝒊𝒆𝒗𝒆 𝑎𝑙𝑙 𝑠𝑢𝑚𝑠 𝑖𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑎𝑠 𝑤𝑒𝑙𝑙
We need to count the number of numbers from 12 to 2, which is:
12 − 2 + 1 = 11

Total
B. 1 2 3 4 5 6
1 2 3 4 5 6 7
2 8
3 9
4 10
5 11
6 12

Example 3.22: Sum of Dice Rolls

Shirley rolls a six-sided dice, with faces numbered from one to six, and then rolls a ten-sided dice, with faces
numbered from one to ten, and adds the numbers on the top face of the two dice. What is the number of distinct
totals that she can get?

{1,2,3,4,5,6}
⏟ , {1,2,3,
⏟ … ,9,10} ⇒ 𝑀𝑖𝑛 𝑆𝑢𝑚 = 1 + 1 = 2, 𝑀𝑎𝑥 𝑆𝑢𝑚 = 6 + 10 = 16
𝑭𝒊𝒓𝒔𝒕 𝑫𝒊𝒄𝒆 𝑺𝒆𝒄𝒐𝒏𝒅 𝑫𝒊𝒄𝒆
𝑊𝑒 𝑐𝑎𝑛 𝑔𝑒𝑡 𝑎𝑙𝑙 𝑠𝑢𝑚𝑠 𝑖𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑎𝑠 𝑤𝑒𝑙𝑙
We need to count the number of numbers from 2 to 16, which is:
16 − 2 + 1 = 15

1 1 1+1=2
2 2
3 3
4 4
5 5
6 6
7
8
9
10 10 + 6 = 16

Example 3.23
Shilpa rolls three six-sided dice, each with faces numbered one to six. What is the number of possible totals that
she can get?

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𝑀𝑎𝑥 = 6 × 3 = 18
𝑀𝑖𝑛 = 1 × 3 = 3
𝑅𝑎𝑛𝑔𝑒 = 18 − 3 + 1 = 16

Challenge 3.24
Shilpi has four dice. The first dice is tetrahedral, with faces numbered from one to four. The second dice is
octahedral, with faces numbered from one to eight. The third dice is dodecahedral, with faces numbered from
one to twelve. The fourth dice is icosahedral, with faces numbered from one to twenty.
She picks anywhere from two to four dice, at random, from the four, and rolls them. The number of totals that
she can get on the rolls is X. What is the difference between the minimum and maximum value of X.

The smallest number of totals is when she picks the tetrahedral and octahedral dice:
𝑀𝑖𝑛 𝑋 = 12 − 2 + 1 = 11
If she picks all four dice:

𝑀𝑎𝑥 𝑋 = 44 − 4 + 1 = 41
𝐷𝑖𝑓𝑓 = 41 − 11 = 30

Example 3.25
I have a spinner with the numbers 2,5 and 8 on it. I spin the spinner three times. What is the number of possible
totals that I can get?

The minimum value that you can get is:

2+2+2=6

The maximum value that you can get is:

8 + 8 + 8 = 24

Notice that the numbers on the spinner have a difference of 3 between them:
8−5 = 5−2= 3

Hence, every time you replace a smaller number with a larger number, you get a total which increases by 3. For
example:
2+2+2=6
2+2+5=9
2 + 2 + 8 = 12
.
.
.
5 + 5 + 8 = 18
5 + 8 + 8 = 21
8 + 8 + 8 = 24

Hence, the possible totals are:

{6, 9,12, … ,24}
3 × {2, 3,4, … ,8}
3 × ({1,2,3, … ,7} + 1)
7 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

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Example 3.26
Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from
each bag, how many different values are possible for the sum of the two numbers on the chips? (AMC 8 2011/8)

3 1+2
5 1+4 3+2
7 1+6 3+4 5+2
9 3+6 5+4
11 5+6
5 Values

𝑂+𝐸 =𝑂

Example 3.27
Bag A has four chips labeled 1, 3, and 5 and 7. Bag B has three chips labeled 2, 4, and 6 and 8. If one chip is
drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?
(AMC 8 2011/8, Adapted)

3 1+2
5 1+4 3+2
7 1+6 3+4 5+2
9 1+8 3+6 5+4 7+2
11 3+8 5+6 7+4
13 5+8 7+6
15 7+8

7 Values

G. Counting on the Number Line

Example 3.28: Counting Natural Numbers on the Number Line

5
How many natural numbers lie between 3 and 2𝜋? (AMC 8 2000/3)

We want to find the natural numbers in the interval:

5
( , 2𝜋)
3
5
Write the numbers in a form which is easier to compare to integers. Write 3 as a mixed number. Write the
decimal value of 2𝜋:
2
(1 , 6.28)
3
Now, it is easier to see that the smallest number that fit the condition is 2, and the largest number is 6:
𝑛{2, … .6} ⇒ 6 − 2 + 1 = 5

Alternate Solution
Another way of arriving at the same solution using a little more mathematical notation is:

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2
⌊2𝜋⌋
⏟ − ⌈1 ⌉ +1 = 6−2+1= 5
⏟ 3
𝑭𝒍𝒐𝒐𝒓 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏
𝑪𝒆𝒊𝒍𝒊𝒏𝒈 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏
The floor function gives us the first number that we will encounter on the left of 2𝜋 on the number line.
2
The ceiling functions gives the first number that we will encounter to the right of 1 on the number line.
3

Example 3.29: Counting Natural Numbers on the Number Line

The number 𝑒 ≈ 2.71 and 𝜋 ≈ 3.14. A frog starts on the real number line at 𝜋. At each hop it jumps to the
natural number that is the nearest to its right. How many hops must it make before the number 3𝑒 lies to the
left of the frog.

3.14 < 𝑥 < 8.13 ⇒ 4,5,6,7,8,9 ⇒ 8 − 4 + 1 = 5

H. Multiple Lists
If we have multiple lists, each following a different set of rules, then we need to break the problem, and calculate
the number of elements of each list separately.

Example 3.30: Counting Digits

Count the number of 𝑑𝑖𝑔𝑖𝑡𝑠
⏟ used to write the numbers from:
𝒏𝒐𝒕 𝑵𝒖𝒎𝒃𝒆𝒓𝒔
A. 1 to 9
B. 10 to 99
C. 100 to 999
D. 1 to 1000
Part A
We have nine numbers. Each number has one digit. Total number of digits is
1×9=9

Part B
We have ninety numbers. Each number has two digits. Total number of digits is
2 × 90 = 180

Part C
We have nine hundred numbers. Each number has three digits. Total number of digits is
3 × 900 = 2700

Part D
= ⏟
9 + 180
⏟ + ⏟
2700 + ⏟
4 = 2893
𝑂𝑛𝑒 𝐷𝑖𝑔𝑖𝑡 𝑇𝑤𝑜 𝐷𝑖𝑔𝑖𝑡 𝑇ℎ𝑟𝑒𝑒 𝐷𝑖𝑔𝑖𝑡 𝐹𝑜𝑢𝑟 𝐷𝑖𝑔𝑖𝑡
𝑁𝑢𝑚𝑏𝑒𝑟𝑠 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 𝑁𝑢𝑚𝑏𝑒𝑟

Example 3.31: Applying the Concept

Prekshak numbered houses from 35 to 784 using wooden digits bought at Rs. 2 per digit. Find the cost.

Prekshak numbered houses from

35,36, … 99,100,101, … 784
We can classify the numbers into:
𝑇𝑤𝑜 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: (35 𝑡𝑜 99) ⇒ 2 𝑑𝑖𝑔𝑖𝑡𝑠 𝑖𝑛 𝑒𝑎𝑐ℎ 𝑛𝑢𝑚𝑏𝑒𝑟

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𝑇ℎ𝑟𝑒𝑒 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: (100 𝑡𝑜 784) ⇒ 3 𝑑𝑖𝑔𝑖𝑡𝑠 𝑖𝑛 𝑒𝑎𝑐ℎ 𝑛𝑢𝑚𝑏𝑒𝑟

⏟
2 (99 − 35 + 1) +
×⏟ ⏟
3 (784 − 100 + 1) = 2 × 65 + 3 × 685 = 130 + 2055 = 2185 𝐷𝑖𝑔𝑖𝑡𝑠
×⏟
𝑫𝒊𝒈𝒊𝒕𝒔 𝑵𝒐.𝒐𝒇 𝑵𝒖𝒎𝒃𝒆𝒓𝒔 𝑫𝒊𝒈𝒊𝒕𝒔 𝑵𝒐.𝒐𝒇 𝑵𝒖𝒎𝒃𝒆𝒓𝒔
𝐶𝑜𝑠𝑡 = ⏟
2 × ⏟
2185 = 4370
⏟
𝐶𝑜𝑠𝑡 𝑝𝑒𝑟 𝑁𝑜.𝑜𝑓 𝐷𝑖𝑔𝑖𝑡𝑠 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡
𝐷𝑖𝑔𝑖𝑡

Example 3.32
An auditorium with 20 rows of seats has 10 seats in the first row. Each successive row has one more seat than
the previous row. If students taking an exam are permitted to sit in any row, but not next to another student in
that row, then the maximum number of students that can be seated for an exam is (AMC 8 1991/13)

Seat 1 2 3 4 5 6 7 8 9 10 11 No. of
No. Students
S S S S S 10
=5
2
S S S S S S 11
=6
2
11
Note: is not 6, but in this case, we are rounding up the number.
2

Seats 10 11 12 13 14 . . . 28 29
Students 5 6 6 7 7 . . . 14 15

We need to add the following twenty numbers:

𝟓 + 𝟔 + 𝟔 + ⋯ + 𝟏𝟒 + 𝟏𝟒 + 𝟏𝟓

We can make pairs, each adding up to twenty. Hence, we can make ten pairs, giving us a total of:
10 × 20 = 200

Example 3.33
An auditorium with 30 rows of seats has 10 seats in the first row. Each successive row has one more seat than
the previous row. If students taking an exam are permitted to sit in any row, but not next to another student in
that row, what is the maximum number of students that can be seated for an exam? (MathCounts 2001 National
Sprint)

Seat 1 2 3 4 5 6 7 8 9 10 11 No. of
No. Students
S S S S S 10
=5
2
S S S S S S 11
=6
2
11
Note: 2
is not 6, but in this case, we are rounding up the number.

Seats 10 11 12 13 14 . . . 38 39
Students 5 6 6 7 7 . . . 19 20

We need to add the following twenty numbers:

𝟓 + 𝟔 + 𝟔 + ⋯ + 𝟏𝟗 + 𝟏𝟗 + 𝟐𝟎
We can make pairs, each adding up to twenty five. Hence, we can make 15 pairs, giving us a total of:

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15 × 25 = 375
I. Bijection to count Elements
Mapping each element of a set to an element of another set creates a bijection.
It is a very powerful tool in counting problems.

If you wish to count the elements of a set 𝐴 = {𝑎, 𝑏, 𝑐 … }, and it is difficult to count, you can map them to
another set 𝐵 = {1,2, 3 … }, which is easier to count. If both the sets have the same number of elements (that is,
the same cardinality), then counting the elements of one set is sufficient to find the number of elements in both
sets.

𝐴 = {𝑎, 𝑏, 𝑐, … . , 𝑧}
𝐵 = {1,2,3, … .26}

You can make pairs by mapping the first element of set A to the first element of set B:
(𝑎, 1), (𝑏, 2), (𝑐, 3), … , (𝑧, 26)

This is also called 1-1 correspondence.

Example 3.34
Find the number of odd numbers from 1 to 75, not including either 1 or 75.

3,5,7, … ,73
Subtract 1 from each number in the list:
2,4,6, … ,72
Divide each number in the list by 2:
1,2,3, … ,36

Example 3.35
Timothy counted the odd numbers starting from 153 (including 153) and going up to 375 (including 375). How
many numbers did he count?

153, 155, … ,375

Add 1 to each number in the list:
154, 156, … ,376
Divide each number in the list by 2:
77,78, … ,188
Count:
188 − 77 + 1 = 112

Core Concept 3.36: Using Bijection

Find the number of elements in the list:
7.25, 7.75,8.25, … ,22.25,22.75
Shortcut
Shortcuts
7.25, 7.75 , 8.25,8.75
⏟ ⏟ , 9.25,9.75
⏟ , … ⇒ 2(22 − 7 + 1) = 2 × 16 = 32
2 7′ 𝑠 2 8′ 𝑠 2 9′ 𝑠
Bijection Method
Subtract 0.25 from each number. This changes the values, but does not change the number of elements:

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7,7.5, 8,8.5, … ,22.5

Double each number to get rid of the fraction in every alternate number:
14,15, 16,17, … ,45
Apply the formula for counting lists:
45 − 14 + 1 = 32 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
Use the properties of Fractions
Convert each term into a fraction, and then make the fraction improper:
1 3 1 1 3 29 31 33 89 91
7 , 7 , 8 , … ,22 , 22 ⇔ , , …, ,
⏟4 4 4 4 4 ⏟
4 4 4 4 4
𝑴𝒊𝒙𝒆𝒅 𝑵𝒖𝒎𝒃𝒆𝒓𝒔 𝑰𝒎𝒑𝒓𝒐𝒑𝒆𝒓 𝑭𝒓𝒂𝒄𝒕𝒊𝒐𝒏𝒔
Multiply each term by 4, to get terms with a difference of 2. Since the terms are not divisible by 2, add 1 to each
term to make it divisible by 2:
29,31,33, … ,89,91 ⇒ 30,32,34, … ,90,92
Divide each term by 2, and then subtract 14 from each term:
15,16,17, … ,45,46 ⇒ 1,2,3, … ,35,32

Example 3.37
Let
𝑋 = {13,15, … . ,85}, 𝑌 = {22,24, … ,92}
Take a number from Set 𝑋. Take a number Set 𝑌. Add the two numbers so obtained to find Z. How many values
can Z take?

𝑀𝑖𝑛 = 13 + 22 = 35
𝑀𝑎𝑥 = 85 + 92 = 177

𝑂+𝐸 =𝑂
All numbers that we can achieve are odd.
All odd numbers in the range 35 to 177 can be achieved.

Hence, we need to count numbers in the list:

𝑛{35,37, … ,177}
Subtract 1 from each number in the list:
𝑛{34,36, … ,176}
Divide each number by 2:
𝑛{17,18, … ,88} ⇒ 88 − 17 + 1 = 72

Example 3.38
A four-digit number can be made by repeating a two-digit number. For example, 1111 is made by repeating 11,
and 1919 is made by repeating 19. How many such numbers are there between 2,000 and 10,000? (Gauss 7
2020/21)

2020, 2121, 2222, …, 9999

Once we choose the first two digits, the last two digits are automatically chosen. Hence, we only focus on
counting the choices for the first two digits.

The smallest value that we can have for the first two digits is 20.
The largest value that we can have is 99

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And we can have all numbers in between, giving us:

20,21,22, . . . ,99 ⇒ 99 − 20 + 1 = 100 − 20 = 80 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

J. Converting numbers into the same form
If the numbers that needed to counted are not in the same format (decimal, percentage, fraction), then it is
convenient to convert all numbers into the same form.
We see an example below

Example 3.39
Find the number of elements that are less than 100 in:
38 44
, 8.2, , 940% ,10, …
5 5

Here, we need to find the last number in the list ourselves.

Method I:
➢ Convert to Fractions with a Denominator of 5
➢ Use the bijection principle
The numbers that we have are in different formats.
38 44
, 8.2
⏟ , , 940%⏟ ,10 …
⏟
5 5 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆
𝑫𝒆𝒄𝒊𝒎𝒂𝒍
𝑭𝒓𝒂𝒄𝒕𝒊𝒐𝒏
Convert everything to fractions to see the pattern better:
38 41 44 47 50
, , , , ,…
5 ⏟
5 5 ⏟
5 5
𝟏 𝟒𝟏 𝟐 𝟒𝟕
𝟖.𝟐=𝟖+𝟎.𝟐=𝟖+ = 𝟗𝟒𝟎%=𝟗.𝟒=𝟗+𝟎.𝟒=𝟗+ =
𝟓 𝟓 𝟓 𝟓
All the denominators are now the same, making them easy to compare. Each numerator is one less than a
500
multiple of three. 100 = 5
fits the pattern, but is not to be included in the counting.
497
Add as the last number in the list.
5
38 41 44 47 50 497
, , , , ,…,
5 5 5 5 5 5

Fractions are much harder to deal with. We don’t want fractions. Multiply each number in the list by 5 to get rid
of the denominator:
38,
⏟ 41 , ⏟
44, 47 , 50 … ,497 ⇒ ⏟
39, 42, 45, … , 498 ⇒ ⏟13, 14, 15 … 166 ⇒ ⏟
1, 2, … ,154 ⇒ 154
⏟
+𝟑 +𝟑 𝑨𝒅𝒅 𝟏 𝒕𝒐 𝒎𝒂𝒌𝒆 𝒆𝒂𝒄𝒉 𝑫𝒊𝒗𝒊𝒅𝒆 𝒃𝒚 𝟑 𝑺𝒖𝒃𝒕𝒓𝒂𝒄𝒕 𝟏𝟐 𝑵𝒖𝒎𝒃𝒆𝒓𝒔
𝒏𝒖𝒎𝒃𝒆𝒓 𝒂 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒆 𝒐𝒇 𝟑

38 41 497 12 × 3 + 2 13 × 3 + 2 165 × 3 + 2
, ,…, = , ,…, ⇒ 165 − 12 + 1 = 154
5 5 5 5 5 5

Method II:
➢ Convert to Decimal Fractions
➢ Write the numbers as multiples of 5
38 44 76 82 88 94 100
, 8.2
⏟ , , 940% ⏟ ,10 … ⇒ , , , , …
⏟
5 5 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 ⏟ 10
10 10 ⏟
10 10
𝑫𝒆𝒄𝒊𝒎𝒂𝒍
𝑭𝒓𝒂𝒄𝒕𝒊𝒐𝒏 𝟖𝟐 𝟗𝟒
𝟖.𝟐= 𝟗𝟒𝟎%=𝟗.𝟒=
𝟏𝟎 𝟏𝟎
Successive numerators have a difference of 6, and are four more than a multiple of 6, so write them like that.

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12 × 6 + 4 13 × 6 + 4
, ,…
10 10
1000
The largest number must be less than 100 = 10 . 1000 is more 4 more than a multiple of 6, so it has a valid
1000−6 994
form, but it too large. So, take 10 = 10 = as the largest number in the sequence:
12 × 6 + 4 13 × 6 + 4 165 × 6 + 4
, ,…, ⇒ 165 − 12 + 1 = 154
10 10 10

K. Finding Multiples
Multiples are important in Number Theory. We often want to find a multiple of a number.

Example 3.40: Finding Multiples

Consider the facts given below that
7 × 11 × 13 = 1001
3 × 333 = 999
Using these, or otherwise, find the
A. seventh multiple of 17
B. largest multiple of 13 which is less than 1000
C. Smallest multiple of 7, which is more than 1000
D. Smallest multiple of 3, which is more than 1000.
E. Largest multiple of 17, which is less than 1000.

𝐷𝑖𝑟𝑒𝑐𝑡 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛: 7 × 17 = 119

Shortcut Method
We will illustrate some shortcuts that involve finding a number that you know
➢ is a multiple of the number that you want
➢ is close to the number that you want
And then adding or subtracting as required in order to meet the other conditions.
1001 = 13𝑥 ⇒ 𝐴𝑛𝑠 = 1001 − 13 = 988
1001 = 7𝑥 ⇒ 𝐴𝑛𝑠 = 1001
999 = 3𝑥 ⇒ 𝐴𝑛𝑠 = 999 + 3 = 1002

Standard Method
The standard method for finding the largest multiple, or the smallest multiple, is to divide the larger number by
the smaller:
1000 14
= 58 ⇒ 𝐴𝑛𝑠 = 1000 − 14 = 986
17 17

Strategy: Try to find numbers near the number that you want that can be calculated easily

Challenge 3.41: Finding Multiples

What is the number of three-digit multiples of 11 ending in 7?

The number must be of the form

𝑎𝑏7
Apply the test of divisibility of 11:
𝑎 + 7 − 𝑏 = 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 11
Range (a + 7) = 8 to 16

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Range(b) = 0 to 9
Case I:
𝑎 + 7 − 𝑏 = 0 ⇒ 𝑎 + 7 = 𝑏 ⇒ (𝑎, 𝑏) = (1,8)(2,9)
187, 297
Case II: a + 7 = b + 11 implies a = b + 4
𝑎 + 7 − 𝑏 = 11 ⇒ 𝑎 = 𝑏 + 4 ⇒ (𝑎, 𝑏) = (4,0)(5,1)(6,2)(7,3)(8,4)(9,5)
407, 517, 627, 737, 847, 957

L. Counting Multiples
When counting multiples, a standard strategy is to:
➢ Find the smallest number and the largest number that meets the conditions
➢ Write out these numbers as multiples.
➢ Count the number of multiples using strategies from counting lists.

Example 3.42
How many multiples of 9 lie between 100 and 200?

Find the smallest multiple of 9 greater than 100:

108
⏟
𝑺𝒎𝒂𝒍𝒍𝒆𝒔𝒕
𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒆
Find the largest multiple of 9 smaller than 200:
198
⏟
𝑳𝒂𝒓𝒈𝒆𝒔𝒕
𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒆
Write out a list of multiples that meets the requirements using the above two:
𝑛{𝟐𝟐 × 9,21 × 9, … ,13 × 9, 𝟏𝟐 × 9}
Divide each number in the list by 9:
𝑛{22,21, … ,12} = 22 − 12 + 1 = 11

Example 3.43
How many multiples of
A. 9 lie between 100 and 200?
B. 6 lie between 100 and 200?
C. 7 lie between 100 and 200?
D. 8 lie between 100 and 200?

Part B, C and D
𝑃𝑎𝑟𝑡 𝐵: {102, … ,198} = {6 × 17, … ,6 × 33} ⇒ 33 − 17 + 1 = 17
𝑃𝑎𝑟𝑡 𝐶: {105, … ,196} = {7 × 15, … ,7 × 28} ⇒ 28 − 15 + 1 = 14
𝑃𝑎𝑟𝑡 𝐷: {104, … ,192} = {8 × 13, … ,8 × 24} ⇒ 24 − 13 + 1 = 12

Example 3.44
Find the number of multiples of 16 between 100 and 200.
Listing Method
We first look at the listing method, which you would have already seen before. This requires finding the first
and the last multiple that satisfy the conditions, converting it into a list, manipulating the list, and then finding
the number of elements in the list:

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{112,128, … ,192} = ⏟
⏟ {16 × 7,16 × 8, … 16 × 12} = ⏟
{7,8, … ,12} = ⏟
12 − 7 + 1 = 6
𝑳𝒊𝒔𝒕 𝒕𝒉𝒆 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒆𝒔 𝑾𝒓𝒐𝒕𝒆 𝒕𝒉𝒆𝒎 𝒂𝒔 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒆𝒔 𝑫𝒊𝒗𝒊𝒅𝒆𝒅 𝒃𝒚 𝟏𝟔 𝑪𝒐𝒖𝒏𝒕𝒆𝒅 𝒕𝒉𝒆 𝒍𝒊𝒔𝒕

Floor and Ceiling Functions

Ceiling Function
Instead of finding the smallest multiple manually, we can also divide 100 by 16
100 50 25 1
100 ÷ 16 = = = =6
16 8 4 4
Now, the above tells us that the sixth multiple of 16 will be smaller than 100. Hence, we need the number after
six, which is seven. In other words, we are rounding up.
Mathematically, this is represented using the ceiling function.
100 1
⌈ ⌉ = ⌈6 ⌉ = 7
⏟16 4
1
𝑅𝑜𝑢𝑛𝑑𝑠 𝑢𝑝 6 𝑡𝑜 7
4
Floor Function
Similarly,
200 1
⌊ ⌋ = ⌊12 ⌋ = 12
16 2

M. Counting Multiple Lists

Till now, the examples that we have seen required us to find the number of elements of a single list.
However, it is possible to combine the two lists together. In such a scenario, it is important to separate out the
lists and then find the numbers of elements in each list.

Example 3.45
What is the number of numbers in the sequence:
6,7,10,11,14, 15, . . . , 94, 95, 98 (𝐴𝑂𝑃𝑆 𝐴𝑙𝑐𝑢𝑚𝑢𝑠)

This can be visualised as two arithmetic sequences:

6, 10, 14....
7, 11, 15....

Both the sequences have a difference of 4. We can rewrite them as:

4 + 2, 8 + 2, 12 + 2. . . . . ,92 + 2, 96 + 2
4 + 3, 8 + 3, 12 + 3, . . . . . .92 + 3

We can further rewrite it as:

4 × 𝟏 + 2, 4 × 𝟐 + 2, 4 × 𝟑 + 2. . . . ., 4 × 𝟐𝟑 + 2, 4 × 𝟐𝟒 + 2 ⇒ 24 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
4 × 𝟏 + 3, 4 × 𝟐 + 3, 4 × 𝟑 + 3, . . . . . .4 × 𝟐𝟑 + 3 ⇒ 23 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

Total number of terms is:

24 + 23 = 47
N. Ceiling and Floor Functions (Optional)
The floor function rounds down a number.
The ceiling functions rounds up a number.
𝑚 𝑛 𝑚 𝑛
Multiples of 𝑥 between 𝑛 and 𝑚 = Floor ( ) − Ceiling ( ) + 1 = ⌊ ⌋ − ⌈ ⌉ + 1
𝑥 𝑥 𝑥 𝑥

Example 3.46: Using the Ceiling and the Floor Functions

How many multiples of 9 lie between 100 and 200? Use the formula with ceiling and floor functions to solve the

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question.

200 100 2 1
Multiples of 9 between 100 and 200 = ⌊ ⌋−⌈ ⌉ = ⌊22 ⌋ − ⌈11 ⌉ = 22 − 12 + 1 = 11
9 9 9 9

Example 3.47: Using the Ceiling and the Floor Functions

Find multiples of 1.5 between 100 and 200.
200 100 2 2 1 2
⌊ ⌋−⌈ ⌉ + 1 = ⌊200 × ⌋ − ⌈100 × ⌉ + 1 = ⌊133 ⌋ − ⌈66 ⌉ + 1 = 133 − 67 + 1 = 67
1.5 1.5 3 3 3 3

3.2 Applications of Lists

A. Start Day and End Day

Example 3.48
A. Ajay travelled to Mexico between Monday and Saturday in the same week. How many days did he travel?
B. Alina travelled to Japan. Her trip was from Monday to Saturday. How many days did she travel?

𝑇𝑢𝑒𝑠𝑑𝑎𝑦, 𝑊𝑒𝑑𝑛𝑒𝑠𝑑𝑎𝑦, 𝑇ℎ𝑢𝑟𝑠𝑑𝑎𝑦, 𝐹𝑟𝑖𝑑𝑎𝑦 ⇒ 4 𝐷𝑎𝑦𝑠

𝑀𝑜𝑛𝑑𝑎𝑦, 𝑇𝑢𝑒𝑠𝑑𝑎𝑦, 𝑊𝑒𝑑𝑛𝑒𝑠𝑑𝑎𝑦, 𝑇ℎ𝑢𝑟𝑠𝑑𝑎𝑦, 𝐹𝑟𝑖𝑑𝑎𝑦, 𝑆𝑎𝑡𝑢𝑟𝑑𝑎𝑦 ⇒ 4 𝐷𝑎𝑦𝑠

B. Start Date and End Date

We can use principles from counting lists to count the number of days that an event takes.

Example 3.49
Gauri’s vacation was from 15th June to 29th June. Find the number of days that she had for vacation.

This can be directly converted into a list:

15 𝐽𝑢𝑛𝑒, 16, 17, … ,29 𝐽𝑢𝑛𝑒 ⇒ 29 − 15 + 1 = 30 − 15 = 15

C. Splitting across Months

If you have an event that is split across different months, the standard technique is to count the days in different
months separately.

Example 3.50
Shivam’s vacation was from 17th June to 8th Aug. Find the number of days that he had for vacation.

𝐷𝑎𝑦𝑠 𝑖𝑛 𝐽𝑢𝑛𝑒: 17,18, … ,30 ⇒ 30 − 17 + 1 = 14

𝐷𝑎𝑦𝑠 𝑖𝑛 𝐽𝑢𝑙𝑦 = 31
𝐷𝑎𝑦𝑠 𝑖𝑛 𝐴𝑢𝑔 = 1,2, … 8 ⇒ 8

𝑇𝑜𝑡𝑎𝑙 = 14 + 31 + 8 = 53
D. Figuring out the Day

Example 3.51
In a particular year, my birthday was on the second Wednesday of February of a non-leap year. My sister’s
birthday was on the 3rd Sunday of the next month. What was the date when my sister’s birthday was celebrated?

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Sun Mon Tue Wed Thu Fri Sat

Feb
14
21
28 1 2 3
4
11
18

Example 3.52
If in the month of March, Sunday, Monday and Tuesday occurred five times, then in the month of September
which days will occur five times?

Sun Mon Tue Wed Thu Fri Sat

1 2 3
8
15
22
29 30 31

31 + 30 + 31 + 30 + 31 + 31

Each of the above has four complete weeks, which will not change the day:
3 + 2 + 3 + 2 + 3 + 3 = 16 = 2 𝑊𝑒𝑒𝑘𝑠 + 2 𝐷𝑎𝑦𝑠

September has 30 days

Sun Mon Tue Wed Thu Fri Sat

1 2
8 9
15
22
29 30

E. Positions
We now look at positions in lists. We start with positions in numbers, and then move to positions in
arrangements.
Finding an element: 7th multiple of 9 ⇔ 7th position in the list of multiples of seven.
Finding a position: When we find 729 is what multiple of 3, we are finding the position of 729 in the list of
multiples of three.

3.53: 𝒏𝒕𝒉 Even and Odd Numbers

𝑛𝑡ℎ even natural number = 2𝑛
𝑛𝑡ℎ odd natural number = 2𝑛 − 1

To find the fifth odd number:

1,3,5,7,9 ⇒ 5𝑡ℎ 𝑁𝑢𝑚𝑏𝑒𝑟 𝑖𝑠 9

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5 × 2 − 1 = 10 − 1 = 9

To find the fourth even number:

2,4,6,8 ⇒ 4𝑡ℎ 𝑁𝑢𝑚𝑏𝑒𝑟 𝑖𝑠 8
4𝑡ℎ 𝑁𝑢𝑚𝑏𝑒𝑟 = 4 × 2 = 8

Example 3.54
A. Find the sum of the tenth even natural number and the twentieth odd natural number.
B. Find the sum of the 18𝑡ℎ odd number, and 35𝑡ℎ even number.

2 × 10 + (2 × 20 − 1) = 20 + 39 = 59

18 × 2 − 1
⏟ + 35
⏟×2 = 35 + 70 = 105
18𝑡ℎ 𝑂𝑑𝑑 𝑁𝑢𝑚𝑏𝑒𝑟 35𝑡ℎ 𝐸𝑣𝑒𝑛 𝑁𝑢𝑚𝑏𝑒𝑟

3.55: Back Calculations

Given a number 𝑛, it is in the
𝑛 𝑡ℎ
( ) even number (if even)
2
𝑛 + 1 𝑡ℎ
( ) odd number (if odd)
2

Example 3.56
If 237 is the 𝑛𝑡ℎ odd number, and 238 is the 𝑚𝑡ℎ even number, what is 3𝑚 − 2𝑛?

238 = 119 × 2 = 119𝑡ℎ even number

237 = 119𝑡ℎ odd number
3𝑚 − 2𝑛 = 3 × 119 − 2 × 119 = 119(3 − 2) = 119

Example 3.57
What is the 1000𝑡ℎ :
A. Odd Integer
B. Integer with odd digits
C. Integer with odd number of digits

Part A:
The 1000th odd integer is
1000 × 2 − 1 = 2000 − 1 = 1999

Part B:
1 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 1,3,5,7,9 ⇒ 5 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
2 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 11,13,15,17,19
⏟ ,⏟
31,33,35,37,39 … , ⇒ 5 × 5 = 25
5 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 5 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
3 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 111,113,115,117,119
⏟ ,⏟
31,33,35,37,39 … , ⇒ 5 × 5 × 5 = 125
5 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 5 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

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4 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 ⇒ 54 = 625

5 + 25 + 125 + 625 = 780

5 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 11111, … ,11999 ⇒ 125 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

780 + 125 = 905

13111, … ,13599 ⇒ 25 × 3 = 75 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

13711, ….

Example 3.58
If Jayshri is fifth from the left and seventh from the right in a row of people seated at a table, what is the total
number of people seated?

The main concept here is that Jayshri gets counted both from the left, and from the right. So, while you might
think we should have:
5 + 7 = 12
We actually need to subtract 1.
Looking at diagram below will help visualize.

1𝑠𝑡 2𝑛𝑑 3𝑟𝑑 4𝑡ℎ 5𝑡ℎ

Jayshri
7𝑡ℎ 6𝑡ℎ 5𝑡ℎ 4𝑡ℎ 3𝑟𝑑 2𝑛𝑑 1𝑠𝑡

1 2 3 4 5 6 7 8 9 10 11

Hence, the final answer is:

5 + 7 − 1 = 12 − 1 = 11 𝑝𝑒𝑜𝑝𝑙𝑒

And note that this is very similar to our formula for counting lists.

Concept 3.59
There are one thousand people seated at a table. (It’s a very long table). Rohini is 𝑥 𝑡ℎ from the right, and 𝑦 𝑡ℎ
from the left. What is 𝑥 + 𝑦?

Leftmost Rightmost
𝑦 𝑡ℎ 𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 1 2 𝑦 998 999 1000
𝑥 𝑡ℎ 𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 1000 999 𝑥 3 2 1

Total 1001 1001 … 1001 … 1001 1001 1001

From the table above, we can see that the sum of any left position, and the corresponding right position is:
1001

Hence, the answer, independent of the value of 𝑥 is

1001

Example 3.60
Patrick Jane, the Mentalist, was tracking a murder suspect. He knew the suspect was among a group playing

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throwball, with the participants numbered consecutively, and spaced equally in a circle. When Patrick reached
the group, the participant numbered 5 threw the ball to the participant standing directly opposite him, who was
numbered 19. If Patrick sizes up the group (taking 10 seconds for each person), the process will take 𝑚 minutes
and 𝑠 seconds (where 𝑠 < 60). Find 𝑚 + 𝑠?

Participants numbered from 6 to 18 are in between participants 5 to 19. This is:

18-5=13 people

The other side will also have

13 people

Total is:
13+13+2=28 people

Time taken:
=28*10=280 seconds=4 minutes 40 seconds
m+s=44
F. Exponents

Example 3.61: Perfect Squares and Cubes

Find the number of:
A. Perfect Squares from 50 to 300
B. Perfect Cubes from 100 to 1000

Part A
We don’t want to list the squares. Instead, we find the smallest number that satisfies, and the largest number
that satisfies

64
⏟ , 81, … , 289
⏟ ⇒ {82 , 92 , … 172 }
𝑺𝒎𝒂𝒍𝒍𝒆𝒔𝒕 𝑳𝒂𝒓𝒈𝒆𝒔𝒕
{ 𝑺𝒒𝒖𝒂𝒓𝒆 𝑺𝒒𝒖𝒂𝒓𝒆 }
Take square roots. The number of square roots is the same as the number of squares (bijection principle):
{8,9, … 17} ⇒ ⏟ 17 − 8 + 1 = 10
𝑪𝒐𝒖𝒏𝒕 𝒕𝒉𝒆 𝑵𝒖𝒎𝒃𝒆𝒓𝒔

Part B
We follow a similar process with finding the number of cubes:

{ 125
⏟ , 216, … ⏟
1000 } ⇒ {53 , 63 , … 93 , 103 }
𝑺𝒎𝒂𝒍𝒍𝒆𝒔𝒕 𝑪𝒖𝒃𝒆 𝑳𝒂𝒓𝒈𝒆𝒔𝒕 𝑪𝒖𝒃𝒆

Take the cube roots, and then count the number of elements in the list:
{5,6, … ,9,10} ⇒ 10 − 5 + 1 = 6

If you take a square root of a perfect square, then you will get a nonnegative number.

√𝑥 < √𝑥 + 1

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3.62: Perfect Square and Perfect Cube

A number is a perfect square and a perfect 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 it is a perfect sixth power.

𝑛6 = (𝑛3 )2 = (𝑛2 )3

Example 3.63: Perfect Squares and Cubes

The sequence 2, 3, 5,6,7,10,11, … contains all the positive integers from least to greatest that are neither squares
nor cubes. What is the 400𝑡ℎ term of the sequence? (MathCounts 2007 Workout 2)

We use complementary counting. We count the numbers that we do not want, and add that numbers at the end
of the list.
𝑃𝑒𝑟𝑓𝑒𝑐𝑡 𝐶𝑢𝑏𝑒𝑠: 13 = 1, 23 = 8, … , 73 = 343, …
𝑃𝑒𝑟𝑓𝑒𝑐𝑡 𝑆𝑞𝑢𝑎𝑟𝑒: 12 = 1, 22 = 4, … , 202 = 400, …

But there are some numbers which are both perfect squares and perfect cubes, which we need to subtract:
16 = 1, 26 = 64 ⇒ 2 𝑛𝑢𝑚𝑏𝑒𝑟𝑠
The final answer is:
400 + 7 + 20 − 2 = 425
As a check, note that:
212 = 441 > 425

Example 3.64
Identify the conditions when the equality
√𝑥 2 = 𝑥
Holds.

𝑥 = 5 ⇒ √52 = 5 ⇒ √𝑥 2 = 𝑥
𝑥 = −5 ⇒ √(−5)2 = 5 ⇒ √𝑥 2 = −𝑥

The above equality is true for 𝑛𝑜𝑛 − 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 numbers.

Example 3.65
How many whole numbers are between √8 and √80? (AMC 8 1986/7)

Whole numbers can also be called non-negative integers.

√8 < 𝑥 < √80 ⇒ 𝑥 ∈ {√9, √16, … √64} ⇒ 𝑥 ∈ {3,4,5,6,7,8} ⇒ 6 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

Example 3.66
How many natural numbers lie between the square root of the largest two-digit number, and the square root of
the smallest four-digit number?

√99 < 𝑥 < √1000 ⇒ √99 < {10,11, … ,31} < √1000 ⇒ 31 − 10 + 1 = 22

Example 3.67
𝑝
How many between √8 and √80 are of the form 3, where 𝑝 is a natural number? (AMC 8 1986/7, Adapted)

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√8 < 𝑥 < √80

√8 = 2√2 ≈ 2 × 1.41 = 2.82

𝑝
Now, we need to identify the smallest number of the form 3 that is greater than 2.82:
8 2 9
= 2 = 2. 6̅ < 2.82 < 3 =
3 3 3
𝑝
And, we also need to identify the largest number of the form that is smaller than √80:
3
27 26
√80 ≈ √81 = 9 = > 8. 6̅
3 3

Hence, we are looking for the number of numbers in the list:

9 10 26
, ,…, ⇒ 9,10, … ,26 ⇒ 26 − 9 + 1 = 27 − 9 = 18
3 3 3

Example 3.68
3 3
How many whole numbers are between √9 and √999?

3 3
√9 < 𝑥 < √999
3 3
√8 = 2, √1000 = 10

We are looking for:

3 3 3
√27, √64, … , √729 ⇒ 3,4,5,6,7,8,9 ⇒ 7 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

Example 3.69
For how many integer values of 𝑏 is 1 ≤ 2𝑏 ≤ 1000

{1,2,4, … ,512} = {20 , 21 , 22 , . . , 29 } ⇒ {0,1,2, … 9} ⇒ 10 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

Example 3.70
For how many integer values of b is 1 < (−2)𝑏 < 1000

This will have the same answer as the one above, except that:
➢ When 𝑏 is even, (−2)𝑏 is positive.
➢ When 𝑏 is odd, (−2)𝑏 is negative.

{1, −𝟐, 4, −𝟖, … ,512} = {20 , 𝟐𝟏 , 22 , 𝟐𝟑 . . , 29 } ⇒ {0,2,4,6,8} ⇒ 5 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

Example 3.71
1
For how many integer values of b is 1000 < 2𝑏 < 1

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1 1 1
{1, , , … , } = {20 , 2−1 , 2−2 , . . , 2−9 } ⇒ {0, −1, −2, … , −9} ⇒ 10 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
2 4 512

Example 3.72
1
For how many integer values of b is 1000 < (−2)𝑏 < 1

G. Number Arrangements

Square
Triangular

∎
∎∎
∎∎∎
∎∎∎∎

If you arrange dots in rows, as above, with an increasing number of dots in each row, you get the shape of a
triangle. Hence, the total number of dots in the arrangement is called a triangular number.
𝑛(𝑛 + 1)
∴ 𝑛𝑡ℎ 𝑇𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑁𝑢𝑚𝑏𝑒𝑟 = 1 + 2 + 3 + ⋯ + 𝑛 =
2

Example 3.73
*Lists and Positions*
What number is directly above $142$ in this array of numbers?

\[\begin{array}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\
\end{array}\]
(AMC 8 1993/24)
Ans = 120

H. Challenge: Partition of Sets

Example 3.74
There are many ways in which the list 0,1,2,3,4,5,6,7,8,9 can be separated into groups. For example, this list
could be separated into the four groups 0,3,4,8 and 1,2,7 and 6 and 5,9. The sum of the numbers in each of these
four groups is 15,10,6, and 14, respectively. In how many ways can the list 0,1,2,3,4,5,6,7,8,9 be separated into at
least two groups so that the sum of the numbers in each group is the same? (Gauss 8 2019/24)

Find the total: 45

If we divide into two groups, then the sum of each group must be:
45/2, not possible

In general, we will only be able to divide into groups which are factors of 45.
Factors of 45 = 1,3,5,9,15,45

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15 groups and above not possible, since we only have 9 numbers

45/9 = 5 Total
This is not possible, since the group with the number 9 will have total at least 9.

45/5 = 9 Total
(9)(1,8)(2,7)(3,6,)(4,5)
0 does not contribute.
So it can go in any group.
So, there are 5 ways to do it.

45/3 = 15 Total
9 has to be in one of the groups
Remaining numbers have to add up to 6.
6
5+1
4+2
1+2+3

(9,6)
Left: 1,2,3,4,5,7,8
To get total of 15:
8+7
8+5+2
8+4+3

(9,6)(8,7)(1,2,3,4,5)
(9,6)(8,5,2)(1,3,4,7)
(9,6)(8,4,3)(1,2,5,7)
(9,6)(8,4,2,1)(3,5,7)
4 Ways

(9,5,1)
Left: 2,3,4,6,7,8

(9,5,1)(7,8)(2,3,4,6)
(9,5,1)(3,4,8)(2,7,6)
2 Ways

(9,4,2)
Left: 1,3,5,6,7,8

(9,4,2)(7,8)(1,3,5,6)
(9,4,2)(1,6,8)(3,5,7)
2 Ways

(9,3,2,1)
Left: 4,5,6,7,8

(9,3,2,1)(7,8)(4,5,6)
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1 Way

4+2+2+1=9
9*3=27

27+5=32

3.3 Visualizing Sets

A. Background
There are many different ways of thinking about sets. There are algebra-heavy methods that work, but may not
be the easiest to do for a question. We look at a few different methods that visualizing sets in different ways. For
an exam questions, the choice of method can be important in deciding the ease with which you can solve the
question, and the time taken.
So, pay careful attention to when which method is preferable. This will be helpful in exam scenarios.

B. Contingency Tables
Contingency tables are a method of presenting information. A simple two-way contingency table classifies
something on the basis of two parameters.
Contingency tables can help us to understand information more quickly. They are also in calculating missing
values, which are much more difficult to calculate otherwise.

Basics 3.75: Creating a Contingency Table

A doctor is considering 50 patients on the basis of whether they have Heart Disease and/or Diabetes. It is also
possible that a patient has neither. The following information is available:
A. 35 have Diabetes
B. 32 have heart disease
C. 18 do not have heart disease
D. 15 do not have Diabetes
E. 20 have both heart disease and Diabetes
F. 12 have heart disease, but not Diabetes
G. 15 have Diabetes, but not heart disease
H. 3 have neither heart disease, nor Diabetes
Present this information in the form of a table.

Diabetes
Yes No Total
Heart Yes 20 12 32
Disease No 15 3 18
Total 35 15 50

Basics 3.76: Interpreting a contingency table

Write the information in the table alongside on people visiting different Visited Europe
countries as points. Yes No Total
Visited Yes 8 7 15
➢ Total Number of People: 80 USA No 33 32 65
➢ Visited Europe: 41 Total 41 39 80
➢ Visited USA: 15

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➢ Not Visited Europe: 39

➢ Not Visited USA: 65
➢ Visited both Europe and USA: 8
➢ Visited Europe, but not USA: 33
VE NVE Total
➢ Visited USA, but not Europe: 7
VUS 8 7 15
➢ Visited neither Europe nor USA: 32
NUS 33 32 65
Total 41 39 80

Example 3.77
I have fourteen children in my block. Five of the children play soccer. Seven of the children play baseball. Three
of the children play both soccer and baseball. Show this information in a two-way contingency table and fill the
rest of the table.

We have information on two things:

➢ Soccer
➢ Baseball
We can put Soccer in the columns and baseball in the rows.
Let 𝑆 = 𝑆𝑜𝑐𝑐𝑒𝑟, 𝑁𝑆 = 𝑁𝑜 𝑆𝑜𝑐𝑐𝑒𝑟, 𝐵 = 𝐵𝑎𝑠𝑒𝑏𝑎𝑙𝑙, 𝑁𝐵 = 𝑁𝑜 𝐵𝑎𝑠𝑒𝑏𝑎𝑙𝑙

S NS Total
B 3 7
NB
Total 5 14

S NS Total
B 3 𝟒=𝟕−𝟑 7
NB 𝟐=𝟓−𝟑 𝟓=𝟗−𝟒 𝟕 = 𝟏𝟒 − 𝟕
𝟓=𝟕−𝟐
Total 5 9 14

S NS Total
B 3 4 7
NB 2 5 7
Total 5 9 14

Self-Check in Contingency Tables

Playing neither soccer, nor baseball, is calculated twice. The number must be the same from both calculations.
In this way, the table is self-checking. If the number is not the same from both the calculations, then:
➢ It is time to go back and check the calculations, and ensure that you have read the question correctly.
➢ It is also possible that the question requires you to establish a contradiction
➢ Finally, it is possible that the question has a mistake.

Example 3.78
There are 40 students in my class. 12 of them learn French, and 17 of them learn Spanish. If 5 of them learn
both French and Spanish, how many of them learn neither of the two languages.

Filling up the information in the question:

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S NS Total
F 5 12
NF
Total 17 40

Filling in the rest of the table:

Spanish
Yes No Total
French Yes 5 7 12
No 12 16 28
Total 17 23 40

The number of people who speak neither of the two languages is 16.

Challenge 3.79
The Pythagoras High School band has 100 female and 80 male members. The Pythagoras High School orchestra
has 80 female and 100 male members. There are 60 females who are members in both band and orchestra.
Altogether, there are 230 students who are in either band or orchestra or both. The number of males in the
band who are NOT in the orchestra is: (AMC 8 1991/23)

Create a contingency table, and add the information in the question:

Band Orchestra Both Total

Female 100 80 60
Male 80 100
Total 180 180 230

Band Orchestra Both Total

Female 100 80 60
Male 80 100 70
Total 180 180 130 230

The number of males in the band who are not in the orchestra
= 80 − 70 = 10

C. Applications of Contingency Tables

Example 3.80: Percentages

At Annville Junior High School, 30% of the students in the Math Club are in the Science Club, and 80% of the
students in the Science Club are in the Math Club. There are 15 students in the Science Club. How many
students are in the Math Club? (AMC 8 1998/14)

Note that the information that is most directly useful is given towards the end.

Science Club No Science Total

Club
Math Club Step I Step II

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80% 𝑜𝑓 15 = 12 12
× 100 = 40
30
No Math Club
Total Given: 15

You can also do this using a Venn Diagram. The calculations are similar, but the presentation is different.

Example 3.81: Fractions and Percentages

There are 30 cars in my building's parking lot. All of the cars are red or white, and a car can have either 2 doors
1
or 4 doors. 3 of them are red, 50% of them are 4-door, and 8 of them are 2-door and white. How many of the
cars are 4-door and red? (AOPS Alcumus, Counting and Probability, Venn Diagrams)

Start with a contingency table. The total number of cars is 30.

Red White
2 Door
4 Door
30

1
3
of them are red, 50% of them are 4-door, and 8 of them are 2-door and white.
Red White
2 Door 8
4 Door 15
10 30

Now fill in the rest of the table:

Red White
2 Door 7 8 15
4 Door 3 12 15
10 20 30

Example 3.82: Ratios

On my ping-pong team there are four times as many ⏟
𝑟𝑖𝑔ℎ𝑡 − ℎ𝑎𝑛𝑑𝑒𝑑 𝑏𝑜𝑦𝑠
⏟ as ⏟
𝑙𝑒𝑓𝑡 − ℎ𝑎𝑛𝑑𝑒𝑑 boys. Of the
𝑹𝑯 𝑩 𝑳𝑯
students on the team who are left-handed, there are twice as many 𝑔𝑖𝑟𝑙𝑠
⏟ as there are 𝑏𝑜𝑦𝑠
⏟ . Half of the girls who
𝑮 𝑩
are on the team are left-handed. If there are 36 people on the team, how many are right-handed boys? (Assume
no player plays ping-pong equally well with both hands.) (AOPS Alcumus, Counting and Probability, Venn

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Diagrams)

When information is presented in terms of ratios, it is often useful to introduce variables in your Venn Diagrams
or Contingency Tables.

𝐿𝑒𝑡 𝑥 = 𝑁𝑜. 𝑜𝑓 ⏟
𝐿𝑒𝑓𝑡 − 𝐻𝑎𝑛𝑑𝑒𝑑 𝐵𝑜𝑦𝑠
⏟
𝑹𝑯 𝑩

Step I Step II Step III

LH RH LH RH LH RH
B 𝑥 4𝑥 B 𝑥 4𝑥 B 𝑥 4𝑥 5𝑥
G 2𝑥 G 2𝑥 2𝑥 G 2𝑥 2𝑥 4𝑥
36 36 3𝑥 6𝑥 9𝑥 = 36

9𝑥 = 36 ⇒ 𝑥 = 4 ⇒ 4𝑥 = 16

Example 3.83
The number of boys in my school who play the guitar is twice the (non-zero) number of girls who play the
guitar. The number of girls who play the flute is twice the number of boys who do so. Also, the number of girls
who play the guitar is twice the number of girls who play the flute. No one in the school plays two instruments.
A. Find the minimum number of children in the school who can play at least one instrument.
B. If the number of children in the school who play the guitar is twenty-four, find the total number of
children in the school.

Part A
The number of boys in my school who play the guitar is twice the number of girls who play the guitar.

Boys Girls
Flute
Guitar 2𝑔 𝑔

The number of girls who play the flute is twice the number of boys who do so.

Boys Girls
Flute 𝑓 2𝑓
Guitar 2𝑔 𝑔

The number of girls who play the guitar is twice the number of girls who play the flute.

Boys Girls
Flute 𝑓 2𝑓
Guitar 2𝑔 = 8𝑓 𝑔 = 4𝑓

Now we can complete the table

Boys Girls Total
Flute 𝑓 2𝑓 3𝑓
Guitar 8𝑓 4𝑓 12𝑓
Total 9𝑓 6𝑓 15𝑓

The number of students who can play at least one instrument is

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15𝑓

And:
𝑀𝑖𝑛 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝑓 = 1 ⇒ 𝑀𝑖𝑛 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 15𝑓 = 15
Part B
No. of children who play the guitar
= 12𝑓 = 24 ⇒ 𝑓 = 2
And hence the total number of children
= 15𝑓 = 30

Example 3.84: Averages

In a liberal arts college, out of 30 students taking math majors, 17 have decided to take a minor in logic, and 14
have decided to take a minor in philosophy. 10 students have decided to take both minors. The average annual
salary of the students taking:
A. neither of the minors is 100,000/year
B. both minors is 97,000/year
C. exactly one minor is 101,000/year

A. Find the total number of students taking exactly one of the minors
B. Find the positive difference in the total salary of the students taking both minors as compared to the
total salary of students taking neither of the minors.

Philosophy No Philosophy Total

Logic 10 17
No
Logic
Total 14 30

Philosophy No Philosophy Total

Logic 10 7 17
No 4 9 13
Logic
Total 14 16 30

(4 + 7) × 101,000 = 11 × 101,000 = 1,111,000

97000 × 10 − 100000 × 9 = 970,000 − 900,000 = 70,000

D. Tree Diagrams

Example 3.85: Fractions

The number of boys in a school is two-thirds of the total student population. One-fourth of the boys play
football, while the rest play basketball. Two-fifths of the girls play football. Everyone in the school plays exactly
one sport. Every classroom in the school has fifty-five children. All classrooms are fully occupied. The number of
children in the school is a four-digit number. What is the minimum number of children in the school?

Tree Diagram

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The total children are divided into

2
➢ boys (3)
1
➢ girls (3) FB (1/4) =1/6
The boys are further divided into Boys (2/3)
those who play BB (3/4) =1/2
1
➢ Football (4) Total
3
➢ Basketball (4) FB (2/5) =2/15
And the girls are further divided into Girls (1/3)
those who play
2 BB (3/5) =1/5
➢ Football (5)
3
➢ Basketball ( )
5

Contingency Table
Suppose that the total number of children in the school is one whole.

Boys Girls Total

Football
Basketball
Total 2 2 1 1
=1− =
3 3 3

Boys Girls
Football 1 2 1 1 2 2
= × = = × =
4 3 6 3 5 15
Basketball 3 2 1 1 3 1
= × = = × =
4 3 2 3 5 5
Total 2 2 1 1
=1− =
3 3 3

Finding the Answer

Boys Girls
Football 1 2
6 15
Basketball 1 1
2 5
Total 2 1 1
3 3

Finding the Answer

The number of children in the school must be a multiple of each of
{6,15,2,5,3}
And hence the minimum number of children in the school is
𝐿𝐶𝑀(6,15,2,5) = 𝐿𝐶𝑀(6,15) = 30

Every classroom has fifty-five children, and all classrooms are fully occupied. Hence, the number of children

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must be a multiple of 55.

And hence the number of children in the school must be
𝐿𝐶𝑀(30,55) = 330𝑛

And the smallest multiple of 330 greater than 1000 is:

330,660,990, 𝟏𝟑𝟐𝟎

E. Line Diagrams (Minimizing and Maximizing)

Sometimes questions are asked, not in terms of actual values for number of elements in the set, but in terms of
the range that the values can take.
Specifically, questions can ask for the minimum number, or the maximum number that a particular region in a
Venn Diagram, or a cell in a contingency table can take.

Example 3.86
Of the 30 students selected for the MOP(𝑀𝑎𝑡ℎ𝑒𝑚𝑎𝑡𝑖𝑐𝑎𝑙 𝑂𝑙𝑦𝑚𝑝𝑖𝑎𝑑 𝑃𝑟𝑜𝑔𝑟𝑎𝑚), 15 students can speak Mandarin
Chinese, and 7 students can speak Hindi. List the possible values of:
A. The number of students who can speak both languages.
B. The number of students who can speak at least one of the two languages.
C. The number of students who can speak neither of the two languages.

{0,1, … ,7}
{15,16, … ,22}
{8, 9, … , 15}

Example 3.87
Set A has 3 elements and Set B has 6 elements. What can be the minimum number of elements in the set 𝐴 ∪ 𝐵?
(JEE Advanced, 1980)

To determine the minimum, let the overlap between Set A and Set B be
maximum.
𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵) = 3 + 6 − 3 = 6

In the above, what would be the maximum?

To determine the maximum, let the overlap between Set A and
Set B be minimum.
𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵) = 3 + 6 − 0 = 9

Example 3.88
One hundred people were surveyed. Of these, 87 indicated they liked Mozart and 70 indicated they liked Bach.
What is the minimum number of people surveyed who could have said they liked both Mozart and Bach?
(MathCounts 2008 School Countdown)

87 + 70 − 100 = 57

Example 3.89
Twelve students in Mrs. Stephenson's class have brown eyes. Twenty students in the class have a lunch box. Of

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Mrs. Stephenson's 30 students, what is the least possible number of students who have brown eyes and a lunch
box? (MathCounts 2005 Chapter Countdown)

Example 3.90
Ellen baked 2 dozen cupcakes of which half contained chocolate, two-thirds contained raisins, one-fourth
contained chocolate chips, and one-sixth contained nuts. What is the largest possible number of cupcakes that
had none of these ingredients? (MathCounts 2008 School Countdown)

Example 3.91
Eighty percent of adults drink coffee and seventy percent drink tea. What is the smallest possible percent of
adults who drink both coffee and tea? (MathCounts 2007 National Countdown)

𝑀𝑖𝑛 = 70 + 80 − 100 = 50
50

Example 3.92
Three-fourths of the students in Mr. Shearer's class have brown hair and six-sevenths of his students are right-
handed. If Mr. Shearer's class has 28 students, what is the smallest possible number of students that could be
both right-handed and have brown hair? (MathCounts 2004 National Sprint)

17
F. Fractions in Minimum and Maximum

Example 3.93
One-third of my school likes to swim. One-half of my school likes to dance. Find the maximum and the minimum
fraction of people in my school who like neither.

Maximum
To maximize the fraction of students who like neither, the overlap between the fraction of people who swim and
the fraction of people who dance should be maximum.
This is illustrated in the diagram below, where all people who swim, are people who dance.

And hence, fraction of people who like neither is

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1 1
1− =
2 2
Minimum
To minimize the fraction of students who like neither, there should be no overlap between the people who swim
and the people who dance.
This is illustrated in the diagram below, where all people who swim, are people who dance.

Hence, the fraction of people who like neither is:

1 1 1
1− − =
2 3 6

In the question above, if the number of students in the school is 60, what is the
A. Maximum number of students who like neither?
B. Minimum number of students who like neither?

1
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 = 60 × = 30
2
1
𝑀𝑖𝑛𝑖𝑚𝑢𝑚 = 60 × = 10
6

Example 3.94: (Alternate Method)

One-third of my school likes to swim. One-half of my school likes to dance. Find the maximum and the minimum
fraction of people in my school who like neither.
Base Case
Let the total number of people in the school be one whole. We can then fill in the following table:

Base Case
Dance Don’t Total
Dance
Swim 1
3
Don’t 2
Swim 3
Total 1 1 1
2 2
Minimum Value
To minimize the green cell, maximize the cells above and to the left of it.
1 1 1
➢ The maximum that the cell above it will take = 𝑀𝑖𝑛 (3 , 2) = 3
1 2 1
➢ The maximum that the cell to the left of it can take will = 𝑀𝑖𝑛 (2 , 3) = 2

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Minimum Dance Don’t Total

Dance
Swim 0 1 1
3 3
Don’t 1 1 2
Swim 2 6 3
Total 1 1 1
2 2

Maximum Value
To maximize the green cell, minimize the cells above and to the left of it.
1
➢ The minimum that the cell above it will take = 0. This let us put the entire value of 2 in the column for
Don’t Dance in the green cell.
➢ We now validate this by finding the value that cell to the left will take, which is
2 1 1
− =
3 2 6

Maximum Dance Don’t Total

Dance
Swim 1 0 1
3 3
Don’t 1 1 2
Swim 6 2 3
Total 1 1 1
2 2

Example 3.95
2𝑡ℎ 3𝑡ℎ
In a room, 5 of the people are wearing gloves, and 4 of the people are wearing hats. What is the minimum
number of people in the room wearing both a hat and a glove? (AMC 8 2010/20)
Strategy
Our contingency table will have fractions.
Consider the number of people in the room as 1 whole.
Contingency Table
Gloves No Gloves Total
Hats 3
4
No Hats
Total 2 1
5

Two values can be calculated directly;

Gloves No Gloves Total
Hats 3
4
No Hats 1
4
Total 2 3 1
5 5

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And since we want to minimize the number of people wear both a hat and a glove, we maximise the number of
people
➢ with no gloves
➢ with no hats
Gloves No Gloves Total
Hats 3 3 3
20 5 4
No Hats 1 0 1
4 4
Total 2 3 1
5 5
This lets us calculate the minimum number of people with both hats and gloves two ways
3 3 15 12 3
− = − =
4 5 20 20 20
2 1 8 5 3
− = − =
5 4 20 20 20
3
→ 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑝𝑒𝑜𝑝𝑙𝑒 = 3
20

3.4 Venn Diagrams: Two Sets

A. Revision: Basics of Sets
Sets are collections of objects. For example, the set of prime numbers can be written as:
𝑃 = {2,3, 5, 7, 11, 13, 17, 19, 23, … } ⇔ ⏟
⏟ 𝑃 = {𝑥 | 𝑥 𝑖𝑠 𝑎 𝑝𝑟𝑖𝑚𝑒}
𝑹𝒐𝒔𝒕𝒆𝒓 𝑭𝒐𝒓𝒎 𝑺𝒆𝒕 𝑩𝒖𝒊𝒍𝒅𝒆𝒓 𝑭𝒐𝒓𝒎
Some terminology related to sets:
➢ A member of a set is called an element of the set.
➢ The number of elements of a set is called its cardinality. The cardinality of the set S is written 𝑛(𝑆).
➢ The set consisting of all elements not in a set is called the complement of the set.
✓ Let there be a set 𝑆.
✓ The complement of 𝑆 is denoted 𝑆 ′ .
➢ The set consisting of all elements presently under consideration is called the Universal set
✓ The universal set is usually denoted 𝑈.
Terminology:
➢ The set consisting of elements in either of the sets is called union of
the two sets.
✓ Union of Set A and Set B is written 𝐴 ∪ 𝐵
➢ The set consisting of elements that are present in both the sets is
called the intersection of the sets.
✓ Intersection of Set A and Set B is written 𝐴 ∩ 𝐵

B. Union of Two Sets

Questions on Venn Diagrams can be solved by multiple methods. The choice of method is a matter of preference.
However, certain questions can be solved much more easily using a specific method.
Hence, it is good to know (and practice!) multiple methods of arriving at the same answer.

3.96: Union of Two Sets1

1
This same formula is used with slightly different notation in probability.
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)

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𝑛(𝐴 ∪ 𝐵)
⏟ = 𝑛(𝐴)
⏟ + 𝑛(𝐵)
⏟ − 𝑛(𝐴 ∩ 𝐵)
⏟
𝑼𝒏𝒊𝒐𝒏 𝒐𝒇 𝑨 𝒂𝒏𝒅 𝑩 𝑬𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝑨 𝑬𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝑩 𝑬𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝑰𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏
𝒐𝒇 𝑨 𝒂𝒏𝒅 𝑩

Example 3.97
Given that
𝑛(𝐴) = 5, 𝑛(𝐵) = 13, 𝑛(𝐴 ∩ 𝐵) = 3, 𝑛(𝐴 ∪ 𝐵) = 15
Verify that
𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵)

Wrong Method
15 = 5 + 13 − 3
15 = 15
You cannot use the equals sign before you have established equality.

Right Method
Hence, calculate the LHS (𝐿𝑒𝑓𝑡 𝐻𝑎𝑛𝑑 𝑆𝑖𝑑𝑒) and the RHS (𝑅𝑖𝑔ℎ𝑡 𝐻𝑎𝑛𝑑 𝑆𝑖𝑑𝑒) separately
𝐿𝐻𝑆 = 15
𝑅𝐻𝑆 = 5 + 13 − 3 = 15
𝐿𝐻𝑆 = 𝑅𝐻𝑆 = 15

Example 3.98
I have fourteen children in my class. Twelve of the children speak English. Seven of the students speak Hindi. If
everyone speaks at least one language, how many students speak both English and Hindi?

5 𝑠𝑝𝑒𝑎𝑘 𝑏𝑜𝑡ℎ 𝐸𝑛𝑔𝑙𝑖𝑠ℎ 𝑎𝑛𝑑 𝐻𝑖𝑛𝑑𝑖

𝑛(𝐸 ∪ 𝐻) = 𝑛(𝐸) + 𝑛(𝐻) − 𝑛(𝐸 ∩ 𝐻)

14 = 12 + 7 − 𝑛(𝐸 ∩ 𝐻)
𝑛(𝐸 ∩ 𝐻) = 19 − 14 = 5

Eng No Eng Total

Hindi 5 2 7
No 7 0 7
Hindi
Total 12 2 14

Example 3.99
I have fourteen children in my block. Five of the children play soccer. Seven of the children play baseball. Three
of the children play both soccer and baseball. Find the number of children who play at least one of the two
sports.

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At least one of the two sports means children who play

➢ Only Soccer
➢ Only Baseball
➢ Both soccer and Baseball
It does not include children who do not play either of the
sports.

𝑆 = 𝑆𝑒𝑡 𝑜𝑓 𝑆𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑤ℎ𝑜 𝑝𝑙𝑎𝑦 𝑆𝑜𝑐𝑐𝑒𝑟,

𝐵 = 𝑆𝑒𝑡 𝑜𝑓 𝑆𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑤ℎ𝑜 𝑝𝑙𝑎𝑦 𝐵𝑎𝑠𝑒𝑏𝑎𝑙𝑙
𝑛(𝑆) = 5, 𝑛(𝐵) = 7, 𝑛(𝑆 ∩ 𝐵) = 3, 𝑈 = 14

𝑛(𝑆 ∪ 𝐵) = 𝑛(𝑆) + 𝑛(𝐵) − 𝑛(𝑆 ∩ 𝐵) = 5 + 7 − 3 = 9

In further questions, we will try and use letters that are

meaningful. Sets will not be explicitly defined.

We presented the same information in a contingency table earlier. Can the same question be answered using the
contingency table? If so, how?

Method I
If we add up the boxes, we will get the total number of students who play either
S NS Total
baseball or soccer B 3 4 7
NB 2 5 7
⏟
3 + ⏟
4 + ⏟
2 =9 Total 5 9 14
𝑷𝒍𝒂𝒚 𝑩𝒂𝒔𝒆𝒃𝒂𝒍𝒍 𝑷𝒍𝒂𝒚 𝑩𝒂𝒔𝒆𝒃𝒂𝒍𝒍, 𝑷𝒍𝒂𝒚 𝑺𝒐𝒄𝒄𝒆𝒓,
𝒂𝒏𝒅 𝑺𝒐𝒄𝒄𝒆𝒓 𝒃𝒖𝒕 𝒏𝒐𝒕 𝒔𝒐𝒄𝒄𝒆𝒓 𝒃𝒖𝒕 𝒏𝒐𝒕 𝑩𝒂𝒔𝒆𝒃𝒂𝒍𝒍

Method II
If we add subtract the maroon box from the blue box, we get the total number of
S NS Total
students who play either baseball or soccer:
B 3 4 7
14
⏟ − ⏟
5 =9 NB 2 5 7
𝑻𝒐𝒕𝒂𝒍 𝑺𝒕𝒖𝒅𝒆𝒏𝒕𝒔 𝑷𝒍𝒂𝒚 𝒏𝒆𝒊𝒕𝒉𝒆𝒓
𝒔𝒐𝒄𝒄𝒆𝒓 𝒏𝒐𝒓 𝒃𝒂𝒔𝒆𝒃𝒂𝒍𝒍 Total 5 9 14

Method III S NS Total

We subtract the red box from the sum of the blue boxes: B 3 4 7
⏟
5 + ⏟
7 − ⏟
3 =9 NB 2 5 7
𝑷𝒍𝒂𝒚 𝑺𝒐𝒄𝒄𝒆𝒓 𝑷𝒍𝒂𝒚 𝑩𝒂𝒔𝒆𝒃𝒂𝒍𝒍 𝑷𝒍𝒂𝒚 𝒃𝒐𝒕𝒉
Total 5 9 14

Example 3.100
Thirteen students in my class learn Math. Twelve students in my class learn Science. All students in my class
take at least one subject. If the number of students who learn both Math, and Science is five, what is the number
of students:
A. who only learn Maths?
B. who only learn Science?
C. In all in my class
D. Who learn either Math, or Science, but not both.

Venn Diagrams

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Let the blue circle be the people who learn Math, and the red circle be the people who
learn Science.
8 5 7
Number of people who only learn Math:
𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑛 𝑡ℎ𝑒 𝐵𝑙𝑢𝑒 𝑐𝑖𝑟𝑐𝑙𝑒, 𝑏𝑢𝑡 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑅𝑒𝑑 𝐶𝑖𝑟𝑐𝑙𝑒 = 8

Number of people who only learn Science:

𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑅𝑒𝑑 𝑐𝑖𝑟𝑐𝑙𝑒, 𝑏𝑢𝑡 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝐵𝑙𝑢𝑒 𝐶𝑖𝑟𝑐𝑙𝑒 = 7

People in all in my class:

8 + 5 + 7 = 20

People who learn Math or Science, but not both:

⏟
8 + ⏟
7 = 15
𝑂𝑛𝑙𝑦 𝑀𝑎𝑡ℎ 𝑂𝑛𝑙𝑦 𝑆𝑐𝑖𝑒𝑛𝑐𝑒

Algebra
Number of people who only learn Math:
𝑛(𝑀 ∩ 𝑆 ′ ) = 𝑛(𝑀) − 𝑛(𝑀 ∩ 𝑆) = 13 − 5 = 8
Number of people who only learn Science:
𝑛(𝑀′ ∩ 𝑆) = 𝑛(𝑆) − 𝑛(𝑀 ∩ 𝑆) = 12 − 5 = 7

The total number of students in the class:

𝑛(𝑀 ∪ 𝑆) = 𝑛(𝑂𝑛𝑙𝑦 𝑀) + 𝑛(𝑂𝑛𝑙𝑦 𝑆) + 𝑛(𝑀 𝑎𝑛𝑑 𝑆) = 8 + 7 + 5 = 20
𝑛(𝑀 ∪ 𝑆) = 𝑛(𝑀) + 𝑛(𝑆) − 𝑛(𝑀 ∩ 𝑆) = 13 + 12 − 5 = 25 − 5 = 20

Contingency Table
Science No Science Total
Math 5 13
No Math 0
Total 12

Science No Science Total

Math 5 8 13
No Math 7 0 7
Total 12 8 20

Example 3.101
Every student in the senior class is taking history or science and 85 of them are taking both. If there are 106
seniors taking history and 109 seniors taking science, how many students are in the senior class? (MathCounts
2002 State Sprint)

𝑛(𝐻 ∪ 𝑆) = 𝑛(𝐻) + 𝑛(𝑆) − 𝑛(𝐻 ∩ 𝑆) = 106 + 109 − 85 = 130

A. Number Theory
Many questions in number theory are related to counting the union of two sets.

Example 3.102

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How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both? (AMC 8 1991/9)

𝑛{3,6, . . ,45} ⇒ 15
𝑛{5,10, . . ,45} ⇒ 9
𝑛{15,30,45} ⇒ 3

𝑛(3 ∪ 5) = 𝑛(3) + 𝑛(5) − 𝑛(3 ∩ 5) = 15 + 9 − 3 = 21

Example 3.103
How many numbers from 1 to 100 are divisible by 2, or 3, or both?

The numbers divisible by are 2 are:

{2,4, … ,100} ⇒ {1 × 2,2 × 2, … ,50 × 2} ⇒ 50 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

The numbers divisible by are 3 are:

{3,6, … ,99} ⇒ {1 × 3,2 × 3, … ,33 × 3} ⇒ 33 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

The numbers divisible by 𝐿𝐶𝑀(2,3) = 6 are:

{6,12, … ,96} ⇒ {1 × 6,2 × 6, … ,16 × 6} ⇒ 16 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

𝑛(𝑇𝑤𝑜 ∪ 𝑇ℎ𝑟𝑒𝑒) = 𝑛(𝑇𝑤𝑜) + 𝑛(𝑇ℎ𝑟𝑒𝑒) − 𝑛(𝑇𝑤𝑜 ∩ 𝑇ℎ𝑟𝑒𝑒) = 50 + 33 − 16 = 67

Example 3.104
If all multiples of 3 and all multiples of 4 are removed from the list of whole numbers 1 through 100, then how
many whole numbers are left? (MathCounts 1991 State Countdown)

Numbers which are multiples of 3 or 4 are:

33 + 25 − 8 = 50
And after the number are removed, the numbers which are left are:
100 − 50 = 50

Example 3.105
How many numbers between 2000 and 3000 are not multiples of three or four?

Step I: Count the multiples

2999 2001
𝑛(3) = ⌊ ⌋−⌈ ⌉ + 1 = 999 − 667 + 1 = 333
3 3
2999 2001
𝑛(4) = ⌊ ⌋−⌈ ⌉ + 1 = 749 − 501 + 1 = 249
4 4
2999 2001
𝑛(12) = ⌊ ⌋−⌈ ⌉ + 1 = 249 − 167 + 1 = 83
12 12
𝑛(3 ∪ 4) = 𝑛(3) + 𝑛(4) − 𝑛(3 ∩ 4) = 333 + 249 − 83 = 499

Step II: Numbers from 2001 to 2999

2999 − 2001 + 1 = 999

Step III: Complementary Counting

We want the numbers which are not multiples of three and four, so subtract them from 999:
999 − 499 = 500 numbers

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Example 3.106
How many numbers between 100 and 1000 are either 𝑝𝑒𝑟𝑓𝑒𝑐𝑡
⏟ 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 or 𝑝𝑒𝑟𝑓𝑒𝑐𝑡
⏟ 𝑐𝑢𝑏𝑒𝑠?
𝑃𝑆 𝑃𝐶

A number which is a perfect square is of the form

𝑛2 , 𝑛∈ℕ
𝑛(𝑃𝑆) = 𝑛{121, … 961} = 𝑛{11, … 31} = 31 − 11 + 1 = 21

A number which is a perfect cube is of the form

𝑛3 , 𝑛∈ℕ

𝑛(𝑃𝐶) = 𝑛{125, … 729} = 𝑛{5, 6, 7, 8, 9} = 5

And, we need to find the numbers which are both perfect squares and perfect cubes, and hence they are of both
forms
𝑛2 , 𝑛3 ⇒ 𝑛𝐿𝐶𝑀(2,3) = 𝑛6 , 𝑛∈ℕ
𝑛(𝑃𝑆 ∩ 𝑃𝐶) = {729} = 1

𝑛(𝑃𝑆 ∪ 𝑃𝐶) = 𝑛(𝑃𝑆) + 𝑛(𝑃𝐶) − 𝑛(𝑃𝑆 ∩ 𝑃𝐶) = 21 + 5 − 1 = 25

Challenge 3.107
The increasing sequence 2,3,5,6,7,10,11, . .. consists of all positive integers that are neither the square nor the
cube of a positive integer. Find the 500th term of this sequence. (AIME 1990/1)

528

Challenge 3.108
Find between 1/1000 and 1000, how many numbers are integer powers of 2 or 3?

1 1
𝑃(2) = 𝑛 ( , … ,256, 512) = 𝑛(2−9 , 2−8 … , 28 , 29 ) = 9 − (−9) + 1 = 19
512 256
1 1
𝑃(3) = 𝑛 ( , … ,243, 729) = 𝑛(3−6 , 3−5 … , 35 , 36 ) = 6 − (−6) + 1 = 13
729 243
𝑃(2 ∩ 3) = 1 ⇔ 2𝑥 = 3𝑥 𝑖𝑓𝑓 𝑥 = 1
𝑃(2 ∪ 3) = 𝑃(2) + 𝑃(3) − 𝑃(2 ∩ 3) = 19 + 13 − 1

B. Single Set
We do need a separate formula for this. Rather, the focus of the questions can be on counting the number of
elements of a single set, given other information.

3.109: Elements belonging only to a set

𝑛(𝑂𝑛𝑙𝑦 𝐴) = 𝑛(𝐴) − 𝑛(𝐴 ∩ 𝐵)

Example 3.110
Twelve students of the twenty students in my class play basketball. Seven students play both basketball and
football. If every person in my class plays least one sport, what is:

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A. The number of students who play football

B. The number of students who play only football

Venn Diagrams
Let the blue circle be basketball, and the red circle be football.
5 7 8
Algebra
Substitute the known values in the formula for the union of a set:
20
⏟ = 𝑛(𝐹) + 12 ⏟ − ⏟ 7 ⇒ 𝑛(𝐹) = 15
𝑛(𝐹∪𝐵) 𝑛(𝐵) 𝑛(𝐹∩𝐵)

Number of people who only football is given by:

𝑛(𝑜𝑛𝑙𝑦 𝐹) = 15
⏟ − ⏟
7 =8
𝑛(𝐹) 𝑛(𝐹 𝑎𝑛𝑑 𝐵)

Contingency Tables
Fill in the information given in the table below.
BB NBB Total
FB 7
NFB 0
Total 12 20

And then the rest of the table can be filled up:

BB NBB Total
FB 7 8 15
NFB 5 0 5
Total 12 8 20

Example 3.111
Sets 𝐴 and 𝐵, shown in the Venn diagram, have the same number of elements. Their
union has 2007 elements and their intersection has 1001 elements. Find the number of
elements in A. (AMC 8 2007/13)

Use the formula for union of two sets:

𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵)

Substitute the known values, and 𝑛(𝐴) = 𝑛(𝐵):

2007 = 2 × 𝑛(𝐴) − 1001 ⇒ 𝑛(𝐴) = 1504

Example 3.112
Sets 𝐴 and 𝐵, shown in the Venn diagram, are such that the total number of elements in set 𝐴 is twice the total
number of elements in set B. Altogether, there are 3011 elements in the union of 𝐴 and 𝐵, and their intersection
has 1000 elements. What is the total number of elements in set 𝐴? (MathCounts 2011 Chapter Sprint)

Use the formula for union of two sets:

𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵)

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Substitute the known values, and also use the fact that 𝑛(𝐴) = 2 × 𝑛(𝐵)
3011 = 2 × 𝑛(𝐵) + 𝑛(𝐵) − 1000

Solve the above linear equation:

𝑛(𝐵) = 1337
Multiply by 2 to get the value of 𝑛(𝐴)
𝑛(𝐴) = 2𝑛(𝐵) = 2674

Example 3.113
In a town of 351 adults, every adult owns a car, motorcycle, or both. If 331 adults own cars and 45 adults own
motorcycles, how many of the car owners do not own a motorcycle? (AMC 8 2011/6)

Algebra / Venn Diagram Method

𝑛(𝐶 ∪ 𝑀) = 𝑛(𝐶) + 𝑛(𝑀) − 𝑛(𝐶 ∩ 𝑀)
351 = 331 + 45 − 𝑛(𝐶 ∩ 𝑀)
𝑛(𝐶 ∩ 𝑀) = 331 + 45 − 351 = 25
..
Car owners do not own a motorcycle
= 𝑛(𝑂𝑛𝑙𝑦 𝐶) = 𝑛(𝐶) − 𝑛(𝐶 ∩ 𝑀) = 331 − 25 = 306

Contingency Table Method

Car No Car Total
MB 45
No MB 0
Total 331 351

Car No Car Total

MB 45
No MB 306 0 306
Total 331 351

C. Intersection of Two Sets

3.114: Intersection of Two Sets

If in the formula for the union of a set, one element is missing, it can be found by substituting the known values.

The formula for the union of two sets is:

𝑛(𝐴
⏟ ∪ 𝐵) = 𝑛(𝐴)
⏟ + 𝑛(𝐵)
⏟ − 𝑛(𝐴
⏟ ∩ 𝐵)
𝑼𝒏𝒊𝒐𝒏 𝒐𝒇 𝑨 𝒂𝒏𝒅 𝑩 𝑬𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝑨 𝑬𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝑩 𝑬𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝑰𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝑨 𝒂𝒏𝒅 𝑩

Example 3.115
Each of the 39 students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a
cat. Twenty students have a dog and 26 students have a cat. How many students have both a dog and a cat?
(AMC 8 2008/11)

Algebraic Method

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39
⏟ = 20
⏟ + 26
⏟ − 𝑛(𝐶 ∩ 𝐷) ⇒ 𝑛(𝐶 ∩ 𝐷) = 20 + 26 − 39 = 7
𝑛(𝐶∪𝐷) 𝑛(𝐷) 𝑛(𝐶)

Contingency Table

Dog No Dog Total

Cat 26
No Cat 0
Total 20 39

Dog No Dog Total

Cat 7 19 26
No Cat 13 0 13
Total 20 19 39

Example 3.116
In a class of 50 students, 28 participate in MATHCOUNTS, 21 participate in science club, and 6 students
participate in neither. How many students participate in both MATHCOUNTS and science club? (MathCounts
2003 National Countdown)

21+28=49
50-6=44

49-44=5

Example 3.117
Fifty students were surveyed about their participation in hockey and baseball. The results of the survey were:
➢ 33 students played hockey
➢ 24 students played baseball
➢ 8 students played neither hockey nor baseball
How many of the students surveyed played both hockey and baseball? (CEMC 2005 Gauss 8)

15

Example 3.118
There are 200 students enrolled at Memorial Middle School. Seventy of the students are in band and 95 are in
chorus. If only 150 students are in band and/or chorus, how many students are in both band and chorus?
(MathCounts 2009 National Sprint)

𝑛(𝐵 ∪ 𝐶) = 𝑛(𝐵) + 𝑛(𝐶) − 𝑛(𝐵 ∩ 𝐶)

150 = 70 + 95 − 𝑛(𝐵 ∩ 𝐶)
𝑛(𝐵 ∩ 𝐶) = 15

3.119: Complement of a Set

The complement of a set is the set of all elements not in the set.
𝑛(𝐴) + 𝑛(𝐴′) = 𝑛(𝑈)

Rather than proving algebraically, the diagram alongside should help you

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understand.

𝐴 = 𝐵𝑙𝑢𝑒 𝑅𝑒𝑔𝑖𝑜𝑛
𝐴′ = 𝑊ℎ𝑖𝑡𝑒 𝑟𝑒𝑔𝑖𝑜𝑛 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝑏𝑢𝑡 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑏𝑜𝑥

𝐴 + 𝐴′ 𝑎𝑑𝑑 𝑢𝑝 𝑡𝑜 𝑡ℎ𝑒 𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙 𝑠𝑒𝑡

Example 3.120
Recall that the universal set is the set of all elements (numbers) under consideration. Also, recall that the
complement of a 𝑆, written 𝑆′ is the set of all elements not in the set. A universal set 𝑈 with 𝑛(𝑈) = 30 has
subsets 𝑋 and 𝑌 such that 𝑛(𝑋) = 18, and 𝑛(𝑌) = 16. If 𝑛(𝑋 ∪ 𝑌)′ = 2, find 𝑛(𝑋 ∩ 𝑌).

Use the property that 𝑛(𝐴) + 𝑛(𝐴′) = 𝑛(𝑈):

𝑛(𝑋 ∪ 𝑌) + 𝑛(𝑋 ∪ 𝑌)′ = 𝑛(𝑈)
𝑛(𝑋 ∪ 𝑌) + 2 = 30
𝑛(𝑋 ∪ 𝑌) = 28

Use the formula for the union of a set:

𝑛(𝑋) + 𝑛(𝑌) − 𝑛(𝑋 ∩ 𝑌) = 28
18 + 16 − 𝑛(𝑋 ∩ 𝑌) = 28
𝑛(𝑋 ∩ 𝑌) = 6

Example 3.121
Let 𝑆 be the set of the 2005 smallest positive multiples of 4, and let 𝑇 be the set of the 2005 smallest positive
multiples of 6. How many elements are common to 𝑆 and 𝑇? (AMC 10A 2005/22)

List the multiples of 4 and the multiples of 6:

4,8, 𝟏𝟐, 16,20, 𝟐𝟒, 28,32, 𝟑𝟔, 40, …
6, 𝟏𝟐, 18, 𝟐𝟒, …

The numbers which are common to both lists are multiples of 𝐿𝐶𝑀(4,6) = 12
We count the number of multiples of 12, which are in the first 2005 multiples of 4.
Every 3rd multiple of 4 is a multiple of 12:
2005 1
= 668
3 3
Drop the fractional part to get:
668

Example 3.122
In the above example, explain what is wrong with the “solution” below:

Every 2nd multiple of 6 is a multiple of 12.

Hence, the number of multiples of 12 to be found in the list 6, 𝟏𝟐, 18, 𝟐𝟒, … is:
2005
= 167
12

6 is a bigger number than 4.

The first 2005 multiples of 6 will include multiples of 12 that are not among the first 2005 multiples of 4.

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D. Elements outside both Sets

3.123: Complement of a Set

The number of elements not in A, and not in B is those elements which are
in the universal set, but outside of both A and B.
They can also be written
𝑛[(𝐴 ∪ 𝐵)′]

Example 3.124
Out of 26 students in a class, 15 students have gone hiking, 3 students have gone both camping and hiking, and
7 students have only gone camping. How many students have gone:
A. Camping
B. for at least one activity
C. for neither camping nor hiking

Define
𝐶 = 𝐶𝑎𝑚𝑝𝑖𝑛𝑔, 𝐻 = 𝐻𝑖𝑘𝑖𝑛𝑔

Part A
Then, the number of students who have only camping:
𝑛(𝐶) = 𝑛(𝑂𝑛𝑙𝑦 𝐶) + 𝑛(𝐶 ∩ 𝐻) = 7 + 3 = 10

Part B
Number of students who went at least one activity means that number of
activities can 1 or 2, but not 0.
This means we want the union of camping and hiking:
𝑛(𝐶 ∪ 𝐻) = 10 + 15 − 3 = 25 − 3 = 22

Part C
This is the number of people outside both the set:
𝑛[(𝐶 ∪ 𝐻)′] = 𝑛(𝑈) − 𝑛(𝐶 ∪ 𝐻) = 26 − 22 = 4

Example 3.125
Out of 50 programmers in an office, 24 program in Python, while 18 program in Java. If the number of people
who program in both is one-third the number of people who program in neither, what is the number of people
who program in neither?

Algebra

𝑛(𝐽) + 𝑛(𝑃) − 𝑛(𝐽 ∩ 𝑃) + 𝑛(𝐽 ∩ 𝑃)′ = 50

2
24 + 18 + 𝑛(𝐽 ∩ 𝑃)′ = 50
3
2
𝑛(𝐽 ∩ 𝑃)′ = 8
3
𝑛(𝐽 ∩ 𝑃)′ = 12

Contingency Table

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Python No Python
Java 𝑥 18
No Java 3𝑥
24 50

Python No Python
Java 𝑥 18 − 𝑥 18
26 − 3𝑥
No Java 3𝑥
24 26 50

18 − 𝑥 = 26 − 3𝑥 ⇒ 2𝑥 = 8 ⇒ 𝑥 = 4 ⇒ 3𝑥 = 12

Example 3.126
In a group of 30 high school students, 8 take French, 12 take Spanish and 3 take both languages. How many
students of the group take neither French nor Spanish? (MOEMS 1 Olympiad 6)

30-(8+12-3)=30-17=13

Example 3.127
Everyone in a class of 30 students takes math and history. Seven students received an A in history and 13
received an A in math, including four that received an A in both courses. How many students did not receive an
A in any of these two courses? (MathCounts 2003 State Countdown)

30 - (7+13-4)=30-16=14

Example 3.128
In a class of 30 students, exactly 7 have been to Mexico and exactly 11 have been to England. Of these 30
students, 4 have been to both Mexico and England. How many students in this class have not been to Mexico or
England? (CEMC 2009 Gauss 7)

16

Example 3.129
A survey of 120 teachers determined the following:
70 had high blood pressure
40 had heart trouble
20 had both high blood pressure and heart trouble
What percent of the teachers surveyed had neither high blood pressure nor heart trouble? (MathCounts 1991
School Sprint)

25

Example 3.130
Roslyn has ten boxes. Five of the boxes contain pencils, four of the boxes contain pens, and two of the boxes
contain both pens and pencils. How many boxes contain neither pens nor pencils? (MathCounts 2005 Chapter
Sprint)

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Example 3.131
In a class of 30 students, exactly 7 have been to Mexico and exactly 11 have been to England. Of these students,
4 have been to both Mexico and England. How many students in this class have not been to Mexico or England?
(Gauss Grade 7 2009/11)

VE NVE Total
VM 4 3 7
NVM 7 16 23
Total 11 19 30

Example 3.132
Out of 200 people in New York, 150 had used the metro, and 80 had used the bus. Twice as many people had
used both as had used neither. What is the number of people who had used both?

𝑈 = 𝑛(𝑀) + 𝑛(𝐵) − 𝑛(𝑀 ∩ 𝐵) + 𝑛(𝑀 ∪ 𝐵)′

200 = 150 + 80 − 2𝑛 + 𝑛
𝑛 = 150 + 80 − 200 = 30
2𝑛 = 60

Challenge 3.133
A Dungeons and Dragons club with 99 members (including the Dungeon Master (DM)) had the following
participation in the campaigns for the year:
➢ Orc Campaign: 61
➢ Troll Campaign: 51
➢ Get Together: 31
Each of the above figures includes the DM. The Get Together was mandatory for those not in any campaigns.
From the members for whom the Get Together was voluntary, the number of members who attended was
exactly double of those who participated in both campaigns. Find the number of people who did not participate
in any of the campaigns.

𝑛(𝑂) + 𝑛(𝑇) − 𝑛(𝑂 ∩ 𝑇) + 𝑛(𝑂 ∪ 𝑇)′ = 𝑛(𝑈)

60 + 50 − 𝑛(𝑂 ∩ 𝑇) + [30 − 2𝑛(𝑂 ∩ 𝑇)] = 98
140 − 98 = 3𝑛(𝑂 ∩ 𝑇)
𝑛(𝑂 ∩ 𝑇) = 14
30 − 2(𝑂 ∩ 𝑇) = 30 − 28 = 2
E. Exclusive OR
Normally when we use the word OR, we mean to say that we want the union. Which means that elements which
belong to both sets should be counted.

But sometimes, we want to count the number of elements that belong to

𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑠𝑒𝑡. This is called the 𝑒𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒 𝑂𝑅.

3.134: Exclusive OR (XOR)

𝐸𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒 𝑂𝑅(𝐴, 𝐵) = 𝑛(𝑂𝑛𝑙𝑦 𝐴) + 𝑛(𝑂𝑛𝑙𝑦 𝐵)

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Example 3.135
Ten students are taking both algebra and drafting. There are 24 students taking algebra. There are 11 students
who are taking drafting only. How many students are taking algebra or drafting but not both? (MathCounts
2001 Workout 2)

Number of students taking only algebra

= 24 − 10 = 14
Number of students taking drafting only
= 11
Students are taking algebra or drafting but not both
= 14 ⏟ + 11 ⏟ = 25
𝑶𝒏𝒍𝒚 𝑶𝒏𝒍𝒚
𝑨𝒍𝒈𝒆𝒃𝒓𝒂 𝑫𝒓𝒂𝒇𝒕𝒊𝒏𝒈

3.136: Exclusive OR (XOR)

𝐸𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒 𝑂𝑅(𝐴, 𝐵) = 𝑛(𝐴)
⏟ + 𝑛(𝐵)
⏟ − 2𝑛(𝐴 ∩ 𝐵)
⏟
𝑬𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝑨 𝑬𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝑩 𝑬𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝑰𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏
𝒐𝒇 𝑨 𝒂𝒏𝒅 𝑩

In the diagram we want the region shaded green and the region shaded blue.
We do not want the region shaded maroon.

𝑛(𝐴) = 𝐺𝑟𝑒𝑒𝑛 + 𝑀𝑎𝑟𝑜𝑜𝑛 ,

⏟ 𝑛(𝐵) = 𝐵𝑙𝑢𝑒 + 𝑀𝑎𝑟𝑜𝑜𝑛
⏟
𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝑰 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝑰𝑰
Add Equations I and II:
𝑛(𝐴) + 𝑛(𝐵) = 𝐺𝑟𝑒𝑒𝑛 + 𝐵𝑙𝑢𝑒 + 2 ∙ 𝑀𝑎𝑟𝑜𝑜𝑛

Move the 2 ∙ 𝑀𝑎𝑟𝑜𝑜𝑛 to the LHS:

𝑛(𝐴) + 𝑛(𝐵) − 2 ∙ 𝑀𝑎𝑟𝑜𝑜𝑛 = 𝐺𝑟𝑒𝑒𝑛 + 𝐵𝑙𝑢𝑒

But note that 𝑀𝑎𝑟𝑜𝑜𝑛 = 𝑛(𝐴 ∩ 𝐵)

𝑛(𝐴) + 𝑛(𝐵) − 2𝑛(𝐴 ∩ 𝐵) = 𝐺𝑟𝑒𝑒𝑛 + 𝐵𝑙𝑢𝑒

And finally, note that 𝐺𝑟𝑒𝑒𝑛 + 𝐵𝑙𝑢𝑒 = 𝐸𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒 𝑂𝑅(𝐴, 𝐵)

𝑛(𝐴) + 𝑛(𝐵) − 2𝑛(𝐴 ∩ 𝐵) = 𝐸𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒 𝑂𝑅(𝐴, 𝐵)

Example 3.137
Let A be the set of integers from 1 to 1000. Find the number of elements of A that are multiples of:
A. 3, but not 4
B. 4, but not 3
C. Either 3, or 4, but not both

First calculate the multiples of 3, 4 and 𝐿𝐶𝑀(3,4) = 12

999
{3,6, … ,999} ⇒ 𝑀(3) = = 333
3
1000
{4,8, … ,1000} ⇒ 𝑀(4) = = 250
4
996
{12,24, … ,996} ⇒ 𝑀(12) = = 83
12
Then, use the formula:
𝑛(𝑂𝑛𝑙𝑦 𝐴) = 𝑛(𝐴) − 𝑛(𝐴 ∩ 𝐵)

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Part A
𝑀(3, 𝑏𝑢𝑡 𝑛𝑜𝑡 4) = 𝑀(3)
⏟ − 𝑀(12)
⏟ = 333 – 83 = 250
𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒆𝒔 𝒐𝒇 𝟑 𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒆𝒔 𝒐𝒇 𝟑 𝒂𝒏𝒅 𝟒
Part B
𝑀(4, 𝑏𝑢𝑡 𝑛𝑜𝑡 3) = 𝑀(4)
⏟ − 𝑀(12)
⏟ = 250 – 83 = 167
𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒆𝒔 𝒐𝒇 𝟒 𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒆𝒔 𝒐𝒇 𝟑 𝒂𝒏𝒅 𝟒
Part C
𝑀(3, 𝑏𝑢𝑡 𝑛𝑜𝑡 4) + 𝑀(4, 𝑏𝑢𝑡 𝑛𝑜𝑡 3) = 250 + 167 = 417

Calculate the number of elements that are multiple of either 3, or 4, but not both using the direct formula

333 + 250 − 2 × 83 = 417

If A is the set of integers between 1 and 1000, then will there be any change in the answer?

The numbers 1 and 1000 are no longer to be considered.

1 is not a multiple of 3, 4, or 12. Hence, no change there.
1000 is not a multiple of 3 or 12. But it is a multiple of 4. Hence, the number of multiples of 4 reduces by one:
𝑀(4) = 250 − 1

𝑀(3, 𝑏𝑢𝑡 𝑛𝑜𝑡 4) remains the same.

𝑀(4, 𝑏𝑢𝑡 𝑛𝑜𝑡 3) reduces by one to be 166.
𝑀(3 𝑜𝑟 4 𝑏𝑢𝑡 𝑛𝑜𝑡 𝑏𝑜𝑡ℎ) reduces by 1 to be 416

Example 3.138
How many numbers between 1 and 2005 are integer multiples of 3 or 4 but not 12? (AMC 10B 2005/13)

F. Subsets
If one set under consideration is a subset of another set, this can be diagrammed in multiple ways. The choice of
diagram depends on preference. You should be aware of the different ways the diagrams
can be made.

Example 3.139
𝐴, 𝐵 and 𝐶 are circular regions as shown. There are 7 items in circle 𝐶. There are exactly
20 items in 𝐴 and 10 of those items are not in 𝐵. How many items are in 𝐵, but not in 𝐶?
(MathCounts 2011 Chapter Sprint)

The number of elements in 𝐵

= 𝑛(𝐴) − 𝑛(𝐴 ∩ 𝐵′ ) = 20 − 10

The number of elements in 𝐵 but not in C

= 𝑛(𝐵) − 𝑛(𝐶) = 10 − 7 = 3

3.5 Venn Diagrams: Three Sets

A. Basics

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Example 3.140
Three flower beds overlap as shown. Bed 𝐴 has 500 plants, bed 𝐵 has 450
plants, and bed 𝐶 has 350 plants. Beds 𝐴 and 𝐵 share 50 plants, while
beds 𝐴 and 𝐶 share 100. The total number of plants is: (AMC 8 1999/9)

350 + 100 + 250 + 50 + 400 = 1150

B. Number of Elements in the Union

The formula for the three set Venn Diagram is more complex
than the two set Venn Diagram.

3.141: Number of Elements in Three Sets

𝑛(𝐴 ∪ 𝐵 ∪ 𝐶) = ⏟ [𝑛(𝐴 ∩ 𝐵) + 𝑛(𝐵 ∩ 𝐶) + 𝑛(𝐴 ∩ 𝐶)] + ⏟
𝑛(𝐴) + 𝑛(𝐵) + 𝑛(𝐶) − ⏟ 𝑛(𝐴 ∩ 𝐵 ∩ 𝐶)
𝑶𝒏𝒆 𝒂𝒕 𝒂𝒕 𝒂 𝑻𝒊𝒎𝒆 𝑻𝒘𝒐 𝒂𝒕 𝒂 𝑻𝒊𝒎𝒆 𝑻𝒉𝒓𝒆𝒆 𝒂𝒕 𝒂 𝒕𝒊𝒎𝒆

Left Hand Side

= 𝑛(𝐴 ∪ 𝐵 ∪ 𝐶) = 𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 + 𝑓 + 𝑔

Right Hand Side

= (𝑎 + 𝑏 + 𝑒 + 𝑑) + (𝑏 + 𝑐 + 𝑒 + 𝑓) + (𝑑 + 𝑒 + 𝑓 + 𝑔)
− [(𝑏 + 𝑒) + (𝑒 + 𝑓) + (𝑑 + 𝑒) ] + 𝑒
= (𝑎 + 2𝑏 + 𝑐 + 2𝑑 + 3𝑒 + 2𝑓 + 𝑔) − (𝑏 + 3𝑒 + 𝑑 + 𝑓) + 𝑒
=𝑎+𝑏+𝑐+𝑑+𝑒+𝑓
= 𝑅𝐻𝑆

Example 3.142
How many integers between 1 and 280, inclusive, are not divisible by 2, 5 or 7? (MathCounts 2022 State Sprint)

Div by 2 = 140
Div by 5 = 56
Div by 7 = 40
Div by 2 and 5: 28
Div by 2 and 7: 20
Div by 5 and 7: 8
Div by 2, 5 and 7: 4

𝑛(𝐴 ∪ 𝐵 ∪ 𝐶) = 𝑛(2) + 𝑛(5) + 𝑛(7) − [𝑛(2 ∩ 5) + 𝑛(5 ∩ 7) + 𝑛(2 ∩ 7)] + 𝑛(2 ∩ 5 ∩ 7)

= 140 + 56 + 40 − [28 + 8 + 20] + 4
= 96

Example 3.143
The summary of a survey of 100 students listed the following totals:
59 students did math homework
49 students did English homework

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42 students did science homework

20 students did English and science homework
29 students did science and math homework
31 students did math and English homework
12 students did math, science and English homework
How many students did no math, no English and no science homework? (MathCounts 2007 Chapter Sprint)

𝑛(𝑀 ∪ 𝐸 ∪ 𝑆) = 59 + 49 + 42 − (20 + 29 + 31) + 12 = 82

𝑛[(𝑀 ∪ 𝐸 ∪ 𝑆)′] = 𝑛(𝑈) − 𝑛(𝑀 ∪ 𝐸 ∪ 𝑆) = 100 − 82 = 18

Example 3.144
An investigator interviewed 100 students to determine their preferences for the three drinks: milk, coffee and
tea. He reported the following:
➢ 10 students had all the three drinks
➢ 20 had milk and coffee
➢ 30 had coffee and tea
➢ 25 had milk and tea
➢ 12 had milk only
➢ 5 had coffee only
➢ 8 had tea only
How many did not take any of the three drinks? (JEE Advanced, 1978)

Draw a diagram and fill in the information:

𝑛(𝑀 ∩ 𝐶 ∩ 𝑇) = 10

Number of people drinking milk and coffee, but not tea

= 𝑛(𝑀 ∩ 𝐶) − 𝑛(𝑀 ∩ 𝐶 ∩ 𝑇) = 20 − 10 = 10

Number of people drinking coffee and tea, but not milk

= 𝑛(𝐶 ∩ 𝑇) − 𝑛(𝑀 ∩ 𝐶 ∩ 𝑇) = 30 − 10 = 20

Number of people drinking milk and tea, but not coffee

= 𝑛(𝑀 ∩ 𝑇) − 𝑛(𝑀 ∩ 𝐶 ∩ 𝑇) = 25 − 10 = 15

𝑛(𝑂𝑛𝑙𝑦 𝑀) = 12
𝑛(𝑂𝑛𝑙𝑦 𝐶) = 5
𝑛 (𝑂𝑛𝑙𝑦 𝑇) = 8

𝑛(𝑀 ∪ 𝐶 ∪ 𝑇) = 12 + 10 + 5 + 15 + 10 + 20 + 8 = 80
𝑛[(𝑀 ∪ 𝐶 ∪ 𝑇)′] = 𝑛(𝑈) − 𝑛(𝑀 ∪ 𝐶 ∪ 𝑇) = 100 − 80 = 20

Example 3.145
Alexio has 100 cards numbered 1-100, inclusive, and places them in a box. Alexio then chooses a card from the
box at random. What is the probability that the number on the card he chooses is a multiple of 2, 3 or 5?
Express your answer as a common fraction. (MathCounts 2002 State Sprint)

𝑛(𝐴 ∪ 𝐵 ∪ 𝐶) = 𝑛(2) + 𝑛(3) + 𝑛(5) − [𝑛(2 ∩ 3) + 𝑛(2 ∩ 5) + 𝑛(3 ∩ 5)] + 𝑛(2 ∩ 3 ∩ 5)

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= 50 + 33 + 20 − [16 + 10 + 6] + 3
= 74

74
𝑃=
100

C. Number of Elements in the Intersection

3.146: Number of Elements in the Intersection

[𝑛(𝐴 ∩ 𝐵) + 𝑛(𝐵 ∩ 𝐶) + 𝑛(𝐴 ∩ 𝐶)] − 2 ⏟
⏟ 𝑛(𝐴 ∩ 𝐵 ∩ 𝐶)
𝑇𝑤𝑜 𝑎𝑡 𝑎 𝑇𝑖𝑚𝑒 𝑇ℎ𝑟𝑒𝑒 𝑎𝑡 𝑎 𝑡𝑖𝑚𝑒

𝑏+𝑒+𝑑+𝑔

𝑛(𝐴 ∩ 𝐵) + 𝑛(𝐵 ∩ 𝐶) + 𝑛(𝐴 ∩ 𝐶)

= (𝑏 + 𝑒) + (𝑒 + 𝑓) + (𝑑 + 𝑒)
= 𝑏 + 𝑒 + 𝑑 + 3𝑒

𝑛(𝐴 ∩ 𝐵) + 𝑛(𝐵 ∩ 𝐶) + 𝑛(𝐴 ∩ 𝐶) − 𝑛(𝐴 ∩ 𝐵 ∩ 𝐶)

= 𝑏 + 𝑑 + 3𝑒 − 2𝑒
= 𝑏+𝑒+𝑑+𝑔

Example 3.147
Every 12 minutes, Bus 𝐴 completes a trip from 𝑃 to 𝑋 to 𝑆 to 𝑋 to 𝑃. Every 20
minutes, Bus B completes a trip from Q to X to T to X to Q. Every 28 minutes,
Bus C completes a trip from R to X to U to X to R. At 1:00 p.m., Buses A, B and C
depart from 𝑃, 𝑄 and 𝑅, respectively, each driving at a constant speed, and
each turning around instantly at the endpoint of its route. Each bus runs until
11:00 p.m. At how many times between 5:00 p.m. and 10:00 p.m. will two or
more buses arrive at X at the same time? (CEMC Gauss Grade 8 2020/24,
Grade 7 2020/25)

Bus A and B
Times for 𝐴 to reach 𝑋 are every six minutes:
{1: 03, 1: 09, … ,2: 03, … ,3: 03, … ,9: 57}
Times for 𝐵 to reach 𝑋 are every ten minutes:
{1: 05, 1: 15, … ,2: 05, … ,3: 05, … ,9: 55}

𝐴 and 𝐵 reach 𝑋 together are every 𝐿𝐶𝑀(6,10) = 30 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 starting 1: 15:

{1: 15,1: 45, … ,9: 45} ⇒ 5 × 2 = ⏟
10 𝑡𝑖𝑚𝑒𝑠
𝒏(𝑨∩𝑩)

Bus B and C
𝐶 reaches 𝑋 at intervals of fourteen minutes.
{1: 07,1: 21, … }

14𝑛 + 7 > 240 ⇒ 𝑆𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑛 = 17

14(17) + 7 = 245 ⇒ 𝐶 𝑟𝑒𝑎𝑐ℎ𝑒𝑠 𝑋 𝑎𝑡
{5: 05,5: 19,5: 33,5: 47, … }

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Since:
𝐿𝐶𝑀(10,14) = 70
Bus B and C reach X together every 70 minutes after 5: 05:
{5: 05,6: 15,7: 25,8: 35,9: 45} ⇒ ⏟
5 𝑡𝑖𝑚𝑒𝑠
𝒏(𝑩∩𝑪)
Bus A and C
𝐿𝐶𝑀(6,14) = 42
Bus A and C reach X together every 42 minutes after 5: 33:
{5: 33, 6: 15, 6: 57, 7: 39, 8: 21, 9: 03, 9: 45} ⇒ ⏟
7 𝑡𝑖𝑚𝑒𝑠
𝒏(𝑨∩𝑪)
Bus A, B and C
All three buses reach together at:
{6: 15, 9: 45} ⇒ ⏟
2 𝑡𝑖𝑚𝑒𝑠
𝒏(𝑨∩𝑩∩𝑪)

10
⏟ + ⏟
5 + ⏟
7 −2× ⏟
2 = 22 − 4 = 18
𝒏(𝑨∩𝑩) 𝒏(𝑩∩𝑪) 𝒏(𝑨∩𝑪) 𝒏(𝑨∩𝑩∩𝑪)

D. Algebra

Example 3.148
A Gourmet Club asked its 52 members which cuisine they liked:
➢ Clue 1: The number of members who liked Chinese was 5 more than those who liked Thai and double of
those who liked Mexican. The ratio of members who liked Chinese, Thai and Mexican was 6: 5: 3.
➢ Clue 2: 25 members liked Thai
➢ Clue 3: 7 members liked both Chinese and Thai while 6 members liked both Thai and Mexican.
➢ Clue 4: The number of members who liked only Thai was one more than the number of members who
liked Mexican.
➢ Clue 5: The number of members who did not like any of Chinese, Thai and Mexican is 4.

Determine the number of members who liked each cuisine

The ratio of members who liked Chinese, Thai and Mexican was
6: 5: 3 = 6𝑥: 5𝑥: 3𝑥

Also, the number of members who liked Chinese was 5 more than those who liked Thai:
6𝑥 = 5 + 5𝑥
𝑥=5

30
⏟ : 25
⏟ : 15
⏟
𝑪𝒉𝒊𝒏𝒆𝒔𝒆 𝑻𝒉𝒂𝒊 𝑴𝒆𝒙𝒊𝒄𝒂𝒏

Determine the number of members who liked all three cuisines

From Clue 4, the number of members who liked only Thai

= 15
⏟ + 1 = 16
𝑳𝒊𝒌𝒆𝒅 𝑴𝒆𝒙𝒊𝒄𝒂𝒏

Out of the 25 members who liked Thai, the members who also liked
Chinese, or Mexican or both are given by
25 − 16 = 9

And from the diagram:

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𝑛(𝐶 ∩ 𝑇) + 𝑛(𝑇 ∩ 𝑀) − 𝑛(𝐶 ∩ 𝑇 ∩ 𝑀 ) = 9

(7 − 𝑥) + (𝑥) + (6 − 𝑥) = 9
13 − 𝑥 = 9
𝑥 = 4, 7 − 𝑥 = 3, 6−𝑥 =2
Alternate Method
𝑛(𝐶 ∪ 𝑇) + 𝑛(𝑇 ∪ 𝑀) − 𝑛(𝐶 ∩ 𝑇 ∩ 𝑀 ) = 9
7 + 6 − 𝑛(𝐶 ∩ 𝑇 ∩ 𝑀 ) = 9
𝑛(𝐶 ∩ 𝑇 ∩ 𝑀 ) = 4

Determine the number of members who liked at least one of the three cuisines

Number of members who like at least one of Chinese, Thai or Mexican is

𝑛(𝐶 ∪ 𝑇 ∪ 𝑀) = 𝑛(𝑈) − 𝑛(𝐶 ∪ 𝑇 ∪ 𝑀)′ = 52 − 4 = 48
We can now draw an updated diagram, focusing only on the three sets.

Determine the number of members who liked Chinese and Mexican

𝑛(𝐶 ∪ 𝑇 ∪ 𝑀) is given by:

𝑛(𝐶) + 𝑛(𝑇) + 𝑛(𝑀) − 𝑛(𝐶 ∩ 𝑇) − 𝑛(𝑇 ∩ 𝑀) − 𝑛(𝐶 ∩ 𝑀) + 𝑛(𝐶 ∩ 𝑇 ∩ 𝑀)

Substituting the known values:

48 = 30 + 25 + 15 − 7 − 6 − 𝑛(𝐶 ∩ 𝑀) + 4
𝑛(𝐶 ∩ 𝑀) = 13

Determine the number of elements in the other regions in the Venn Diagram.

𝑛(𝐶 ∩ 𝑀 ∩ 𝑇′) = 13 − 4 = 9
𝑛(𝑀 ∩ 𝐶 ′ ∩ 𝑇 ′ ) = 15 − 9 − 4 − 2 = 0
𝑛(𝐶 ∩ 𝑇 ′ ∩ 𝑀′ ) = 30 − 3 − 4 − 9 = 14

Challenge 3.149
Jeremy made a Venn diagram showing the number of students in his class who own
types of pets. There are 32 students in his class. In addition to the information in the
Venn diagram, Jeremy knows half of the students have a dog, 3/8 have a cat, six have
some other pet and five have no pet at all. How many students have all three types of
pets (i.e. they have a cat and a dog as well as some other pet)? (MathCounts 2006
State Sprint)

Students in the lettered regions are:

𝑤 + 𝑥 + 𝑦 + 𝑧 = 32 − 5 − 9 − 10 − 2 = 6
But, students in the region 𝑥 + 𝑦 + 𝑧 are also the same:
𝑥 + 𝑦 + 𝑧 = 16 − 10 = 6 ⇒ 𝑤 = 0
⏟
𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝑰
Therefore,
𝑥 + 𝑧 = 12 − 9 = 3 ,
⏟ 𝑧⏟+ 𝑦 = 6 − 2 = 4
𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝑰𝑰 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝑰𝑰𝑰

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𝑦
⏟= 3 , 𝑥
⏟= 2 ⇒𝑧=1
𝑬𝒒.𝑰−𝑬𝒒 𝑰𝑰 𝑬𝒒.𝑰−𝑬𝒒 𝑰𝑰𝑰

Challenge 3.150
(𝑊𝑎𝑟𝑛𝑖𝑛𝑔: 𝐷𝑖𝑓𝑓𝑖𝑐𝑢𝑙𝑡. 𝐹𝑒𝑒𝑙 𝑓𝑟𝑒𝑒 𝑡𝑜 𝑠𝑘𝑖𝑝 𝑎𝑛𝑑 𝑚𝑜𝑣𝑒 𝑜𝑛)
Problems A, B and C were posed in a mathematical contest. 25 competitors solved at least one of the three.
Amongst those who did not solve A, twice as many solved B as C. The number solving only A was one more than
the number solving A and at least one other. The number solving just A equalled the number solving just B plus
the number solving just C. How many solved just B? (IMO 1966 A/1)

Since the earlier equation is in terms of b and c, find

Set up a Venn Diagram
the value of 𝑑 in terms of 𝑏 and 𝑐:
𝐵
𝐴 𝑏 The number who solved A and at least one of B or C:
𝑎 25 − 𝑎 − 𝑏 − 𝑐 − 𝑑

The number solving only A was one more than the

𝑑 number solving A and at least one other. Hence:
𝑎 = 1 + 25 − 𝑎 − 𝑏 − 𝑐 − 𝑑
𝑐 2𝑎 − 𝑏 − 𝑐 − 𝑑 = 26
𝐶 Solve
Let Substitute the value of 𝑑 from Equation II, and the
𝑎 𝑐𝑜𝑚𝑝𝑒𝑡𝑖𝑡𝑜𝑟𝑠 𝑠𝑜𝑙𝑣𝑒 𝑗𝑢𝑠𝑡 𝐴 value of 𝑎 from Equation I:
𝑏 𝑐𝑜𝑚𝑝𝑒𝑡𝑖𝑡𝑜𝑟𝑠 𝑠𝑜𝑙𝑣𝑒 𝑗𝑢𝑠𝑡 𝐵 2(𝑏 + 𝑐) + 𝑏 + 𝑐 + (𝑏 − 2𝑐) = 26
𝑐 𝑐𝑜𝑚𝑝𝑒𝑡𝑖𝑡𝑜𝑟𝑠 𝑠𝑜𝑙𝑣𝑒 𝑗𝑢𝑠𝑡 𝐶 4𝑏 + 𝑐 = 26
𝑑 𝑐𝑜𝑚𝑝𝑒𝑡𝑖𝑡𝑜𝑟𝑠 𝑠𝑜𝑙𝑣𝑒 𝐵 𝑎𝑛𝑑 𝐶 𝑏𝑢𝑡 𝑛𝑜𝑡 𝐴
Since 𝑐 ≥ 0, 𝑏 ≥ 0, we have a Diophantine equation.
Set up Equations 25
The number solving just A equaled the number 𝑐=1⇒𝑏= ⇒ 𝑁𝑜𝑡 𝑉𝑎𝑙𝑖𝑑
4
solving just B plus the number solving just C:
𝑐 = 2, 𝑏 = 6
𝑎 =𝑏+𝑐
⏟
𝑐 = 6, 𝑏 = 4
𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝑰
Amongst those who did not solve A, twice as many But we also know that:
solved B as C. Hence: 𝑑 = 𝑏 − 2𝑐 ≥ 0 ⇒ 𝑏 ≥ 2𝑐 ⇒ 𝑏 = 6, 𝑐 = 2.
𝑏 + 𝑑 = 2(𝑐 + 𝑑) ⇒ ⏟
𝑑 = 𝑏 − 2c Hence:
𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝑰𝑰 𝑏 = 6

E. Number Theory

Example 3.151:
How many numbers between 1000 to 2000 are:
A. Divisible by 3, 4 and 5
B. Divisible by 3 and 4, but not 5
C. Divisible by 3 and 5, but not 4
D. Divisible by 5 and 4, but not 3
E. Divisible by 3, but not 4 and 5
F. Divisible by 4, but not 3 and 5
G. Divisible by 5 but not 3 and 4

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Example 3.152
How many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5? (AMC 10 2001/25)

𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑒𝑠 𝑜𝑓 3 = { }

𝑛(𝐴 ∪ 𝐵 ∪ 𝐶) = 𝑛(𝐴) + 𝑛(𝐵) + 𝑛(𝐶) − 𝑛(𝐴 ∩ 𝐵) − 𝑛(𝐵 ∩ 𝐶) − 𝑛(𝐴 ∩ 𝐶) + 𝑛(𝐴 ∩ 𝐵 ∩ 𝐶)

Example 3.153
Ashley writes out the first 2017 positive integers. She then underlines any of the 2017 integers that is a multiple
of 2, and then underlines any of the 2017 integers that is a multiple of 3, and then underlines any of the 2017
integers that is a multiple of 5. Finally, Ashley finds the sum of all the integers which have not been underlined.
What is this sum? (Gauss Grade 7 2017/25)

We will use complementary counting. Find the sum of all the numbers from 1 to 2017, and then subtract the
sum of the numbers which have been underlined.

𝑛(𝑛+1)
The sum of the first 𝑛 natural numbers is given by .
Hence:
2
2017 × 2018
1 + 2 + ⋯ + 2017 = = 2,035,153
2

𝑛(𝑆2 ∪ 𝑆3 ∪ 𝑆5 ) = 𝑛(𝑆2 ) + 𝑛(𝑆3 ) + 𝑛(𝑆5 ) − 𝑛(𝑆6 ) − 𝑛(𝑆15 ) − 𝑛(𝑆10 ) + 𝑛(𝑆30 )

1008 × 1009
𝑆2 = 2 + 4 + ⋯ + 2016 = 2(1 + 2 + ⋯ + 1008) = 2 ( ) = 1,017,072
2
672 × 673
𝑆3 = 3 + 6 + ⋯ + 2016 = 3(1 + 2 + ⋯ + 672) = 3 ( ) = 678,384
2
403 × 404
𝑆5 = 5 + 10 + ⋯ + 2015 = 5(1 + 2 + ⋯ + 403) = 5 ( ) = 407,030
2
336 × 337
𝑆6 = 6 + 12 + ⋯ + 2016 = 6(1 + 2 + ⋯ + 336) = 6 ( ) = 339,696
2
134 × 135
𝑆15 = 15 + 30 + ⋯ + 2010 = 15(1 + 2 + ⋯ + 134) = 15 ( ) = 135,675
2
201 × 202
𝑆10 = 10 + 20 + ⋯ + 2010 = 10(1 + 2 + ⋯ + 201) = 10 ( ) = 203,010
2

67 × 68
𝑆15 = 30 + 60 + ⋯ + 2010 = 30(1 + 2 + ⋯ + 67) = 30 ( ) = 68,340
2

𝑛(𝑆2 ∪ 𝑆3 ∪ 𝑆5 ) = 1017072
⏟ + 678,384
⏟ + 407,030
⏟ − 339,696
⏟ − 135,675
⏟ − 203,010
⏟ + 68,340
⏟ = 1,492,445
𝑛(𝑆2 ) 𝑛(𝑆3 ) 𝑛(𝑆5 ) 𝑛(𝑆6 ) 𝑛(𝑆15 ) 𝑛(𝑆10 ) 𝑛(𝑆30 )

2,035,153 − 1,492,445

F. Rules

Example 3.154
If A, B and C are three sets such that 𝐴 ∩ 𝐵 = 𝐴 ∩ 𝐶 and 𝐴 ∪ 𝐵 = 𝐴 ∪ 𝐶, then:
A. 𝐴 = 𝐶
B. 𝐵 = 𝐶
C. 𝐴 ∩ 𝐵 = 𝜙

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D. 𝐴 = 𝐵 (JEE Main, 2009)

𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵 ⇒ 𝑥 ∈ 𝐴 ∩ 𝐵 ⇒ 𝑥 ∈ 𝐴 ∩ 𝐶 ⇒ 𝑥 ∈ 𝐶 ⇒ 𝐵 = 𝐶 ⇒ 𝑂𝑝𝑡𝑖𝑜𝑛 𝐵

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4. COUNTING RULES
4.1 Addition and Multiplication Rules
A. Addition Rule
The addition and multiplication rules are key building blocks in counting problems.

4.1: Addition Rule

Suppose an event 𝑨 can happen in 𝒏 ways. And an event 𝑩 can happen in 𝒎 ways. Then, the number of ways in
which either 𝑨 or 𝑩 can happen is:
𝒏+𝒎

Exactly 𝑜𝑛𝑒 of the events happens.

The recommended keyword to look for is OR, but this is not necessary, since questions can be phrased in a
number of ways.

Example 4.2
What is the number of ways of picking a single element from any one of the sets?
𝐴 = {𝑎, 𝑏, 𝑐, 𝑑, … , 𝑛}, 𝐵 = {1,2,3, … , 𝑚}

{𝑎,
⏟ 𝑏, 𝑐, 𝑑, … , 𝑛} + {1,2,3,
⏟ … , 𝑚} ⇒𝑛+𝑚
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑾𝒂𝒚𝒔 𝒐𝒇 𝒑𝒊𝒄𝒌𝒊𝒏𝒈 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒘𝒂𝒚𝒔 𝒐𝒇 𝒑𝒊𝒄𝒌𝒊𝒏𝒈
𝒐𝒏𝒆 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒇𝒓𝒐𝒎 𝑨=𝒏 𝒐𝒏𝒆 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒇𝒓𝒐𝒎 𝑩=𝒎

Example 4.3
At Euclid Middle School the mathematics teachers are Mrs. Germain, Mr. Newton, and Mrs. Young. There are 11
students in Mrs. Germain's class, 8 students in Mr. Newton's class, and 9 students in Mrs. Young's class taking
the AMC 8 this year. How many mathematics students at Euclid Middle School are taking the contest? (AMC 8
2010/1)

This is a direct application of the addition rule:

11
⏟ + ⏟
8 + ⏟
9 = 28
𝑮𝒆𝒓𝒎𝒂𝒊𝒏′ 𝒔 𝑵𝒆𝒘𝒕𝒐𝒏′ 𝒔 𝒀𝒐𝒖𝒏𝒈′ 𝒔
𝑪𝒍𝒂𝒔𝒔 𝑪𝒍𝒂𝒔𝒔 𝑪𝒍𝒂𝒔𝒔

Example 4.4
How many choices do I have in the following situations?
A. Choose a pair of shoes out of four pairs.
B. Travel from Andheri to Churchgate via one of Metro, Bus, or Car.
C. Choose a captain from a team of eleven people.
D. Go from City A to City B, using one of five roads that go from one city to the other.
E. Choose a fruit from Apple, Watermelon, and Chickoo.
F. Wear either a cap or a bandana, if I have 10 distinct caps, and 5 distinct bandanas
G. Enter a room that has four doors, and five windows, if I can enter via either a door or a window?
H. Assign a single alphanumeric character identifying a book. (An alphanumeric character can be from the
English alphabet, or a digit from the decimal system).

Part A
1 out of the 4 pairs = 4 choices
Part B

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Metro / Bus / Car = 3 choices

Part C
Choosing a captain from a team of 11 people = 11 Choices
Part D
One of 5 Roads = 5 Choices
Part E
Choosing one of three fruits = 3 choices
Part F
Cap = 10 Choices
Bandana = 5 Choices
Total = 10 + 5 = 15
Part G
Door = 4 choices
Window = 5 choices
Door or Window = 9 choices
Part H
Number = 10 choices
Letter = 26 choices
Number or Letter = 36 choices

B. Multiplication Rule

4.5: Multiplication Rule

Suppose 𝑨 can happen in 𝒎 ways. And 𝑩 can happen in 𝒏 ways. The number of ways in which 𝑨 and 𝑩 can
happen is:
𝒎×𝒏

Both events A and B are going to happen. This is in contrast to the addition rule, where 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 of A and B
is going to happen.
Hence, any option for event A can be mixed with any option for event B.

Example 4.6
What is the number of ways of picking a single element from A, and a single element from B?
𝐴 = {𝑎, 𝑏, 𝑐, 𝑑, … , 𝑛}, 𝐵 = {1,2,3, … , 𝑚}

𝑛×𝑚

Example 4.7
How many distinct outfits consisting of a shirt and a pair of jeans can I make from four shirts and three pairs of
jeans?

𝐹𝑖𝑟𝑠𝑡 𝑆ℎ𝑖𝑟𝑡: 3 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑓𝑜𝑟 𝐽𝑒𝑎𝑛𝑠

𝑆𝑒𝑐𝑜𝑛𝑑 𝑆ℎ𝑖𝑟𝑡: 3 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑓𝑜𝑟 𝐽𝑒𝑎𝑛𝑠
𝑇ℎ𝑖𝑟𝑑 𝑆ℎ𝑖𝑟𝑡: 3 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑓𝑜𝑟 𝐽𝑒𝑎𝑛𝑠
𝐹𝑜𝑢𝑟𝑡ℎ 𝑆ℎ𝑖𝑟𝑡: 3 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑓𝑜𝑟 𝐽𝑒𝑎𝑛𝑠

𝑇𝑜𝑡𝑎𝑙 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 = 3 + 3 + 3 + 3 = 3 × 4 = 12

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C. Deciding between Addition and Multiplication

Example 4.8
𝐴𝑛𝑠𝑤𝑒𝑟 𝑒𝑎𝑐ℎ 𝑝𝑎𝑟𝑡 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑒𝑙𝑦
Shivansh has three flavors of ice-cream: Roasted Almond, Bitter Chocolate, and Pistachio. He has four fruits:
apple, passionfruit, jackfruit, and pineapple. How many choices does he have if he is going to eat:
A. One ice-cream and one fruit
B. Either an ice cream or a fruit, but not both

Part A
Shivansh can first choose his fruit (4 choices), and Choices
then choose his ice-cream (3 choices).
This corresponds to choosing one of the blue options
Apple Passionfruit Jackfruit Pineapple
on the chart alongside, and then choosing one of the
green options.
Roasted Roasted Roasted Roasted
Almond Almond Almond Almond
In all, there are:
⏟
3 × ⏟
4 ⇒ 3 × 4 = 12
Bitter Bitter Bitter Bitter
𝑾𝒂𝒚𝒔 𝒐𝒇 𝒑𝒊𝒄𝒌𝒊𝒏𝒈 𝑾𝒂𝒚𝒔 𝒐𝒇 𝒑𝒊𝒄𝒌𝒊𝒏𝒈
𝒐𝒏𝒆 𝒊𝒄𝒆−𝒄𝒓𝒆𝒂𝒎 𝒂 𝒇𝒓𝒖𝒊𝒕=𝟒
Chocolate Chocolate Chocolate Chocolate

Pistachio Pistachio Pistachio Pistachio

Part B
{Roasted
⏟ Almond, Bitter Chocolate, Pistachio} + {Apple,
⏟ Passionfruit, Jackfruit, Pineapple} ⇒ 3 + 4 = 7
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑾𝒂𝒚𝒔 𝒐𝒇 𝒑𝒊𝒄𝒌𝒊𝒏𝒈 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒘𝒂𝒚𝒔 𝒐𝒇 𝒑𝒊𝒄𝒌𝒊𝒏𝒈
𝒐𝒏𝒆 𝒊𝒄𝒆−𝒄𝒓𝒆𝒂𝒎=𝟑 𝒂 𝒇𝒓𝒖𝒊𝒕=𝟒

Example 4.9
If Vedika visits the US, she wants to go to one of four cities: Seattle, New York, Philadelphia, or Denver. If she
decides to visit Europe, she wants to go one of five cities: Rome, Milan, London, Istanbul, or Madrid. How many
choices does she have if she is going to visit:
A. A single city
B. A city in the USA, and a city in Europe, and the order of visiting the cities does not matter.

Part A
{Seattle,
⏟ New York, Philadelphia or Denver} Location
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑾𝒂𝒚𝒔 𝒐𝒇 𝒑𝒊𝒄𝒌𝒊𝒏𝒈
𝒐𝒏𝒆 𝒄𝒊𝒕𝒚 𝒇𝒓𝒐𝒎 𝑼𝑺𝑨=𝟒
+ {Rome,
⏟ Milan, London, Istanbul or Madrid} ⇒ 4 + 5 = 9 US Europe
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒘𝒂𝒚𝒔 𝒐𝒇 𝒑𝒊𝒄𝒌𝒊𝒏𝒈
𝒂 𝒄𝒊𝒕𝒚 𝒇𝒓𝒐𝒎 𝑬𝒖𝒓𝒐𝒑𝒆=𝟓
4 Choices 5 Choices

Part B

Europe
Rome Milan London Istanbul Madrid
USA Seattle (𝑆𝑒𝑎𝑡𝑡𝑙𝑒, 𝑅𝑜𝑚𝑒 ) (𝑆𝑒𝑎𝑡𝑡𝑙𝑒, 𝑀𝑖𝑙𝑎𝑛)
New York
Philadelphia

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Denver

{Seattle, New York, Philadelphia or Denver} × ⏟

⏟ {Rome, Milan, London, Istanbul or Madrid} ⇒ 4 × 5 = 20
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑾𝒂𝒚𝒔 𝒐𝒇 𝒑𝒊𝒄𝒌𝒊𝒏𝒈 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒘𝒂𝒚𝒔 𝒐𝒇 𝒑𝒊𝒄𝒌𝒊𝒏𝒈
𝒐𝒏𝒆 𝒄𝒊𝒕𝒚 𝒇𝒓𝒐𝒎 𝑼𝑺𝑨=𝟒 𝒂 𝒄𝒊𝒕𝒚 𝒇𝒓𝒐𝒎 𝑬𝒖𝒓𝒐𝒑𝒆=𝟓

D. “Nothing” as an option
Not doing anything is sometimes also a valid option. This needs to be considered and added to the number of
valid choices.

4.10: Nothing as an Option

If I must choose from 1 of 𝑛 options, and I also the option to not choose, it is a hidden option, and hence the
number of choices is
𝑛+1

Example 4.11
I need to choose my shoes and my socks for my dinner outing. I have shoes in three colours: black, brown, and
white. I have socks in four colours: gray, blue, yellow and red. In how many ways can I choose if:
A. I always wear shoes and socks.
B. I always wear shoes, but also have the option of not wearing socks.

Part A
I need to choose Sock Color
➢ Shoes Gray Blue Yellow Red
➢ Socks Shoe Black
And I can combine them in whichever way I Color Brown Brown Shoe,
Blue Sock
wish. The table shows the options.
White
Each cell in the table corresponds to a choice
of shoe and a choice of socks. For example, the cell in the second row and second column corresponds to a
brown shoe paired with a blue sock.

The total number of choices is:

⏟
3 × ⏟
4 = 12
𝑁𝑜.𝑜𝑓 𝑁𝑜.𝑜𝑓
𝑆ℎ𝑜𝑒𝑠 𝑆𝑜𝑐𝑘𝑠
Part B
If I also have the option of not wearing
socks, my number of options for socks Sock Color
increases by one, making the final Gray Blue Yellow Red No
answer: Socks
⏟
3 × ⏟
5 = 15 Shoe Black
𝑁𝑜.𝑜𝑓 𝑁𝑜.𝑜𝑓 𝑜𝑝𝑡𝑖𝑜𝑛𝑠 Color Brown Brown Shoe,
𝑆ℎ𝑜𝑒𝑠 𝑓𝑜𝑟 𝑆𝑜𝑐𝑘𝑠 Blue Sock
White

Example 4.12
I need to choose my jacket, and my tie. I have a red jacket, a blue jacket, and a green jacket. I have a polka-dotted
tie, and a striped tie. I also have the option of not wearing one or more of the above two items. In how many
ways can I choose my outfit?

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𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑓𝑜𝑟 𝐽𝑎𝑐𝑘𝑒𝑡: 𝑅𝑒𝑑, 𝐵𝑙𝑢𝑒, 𝐺𝑟𝑒𝑒𝑛, 𝑁𝑜 𝐽𝑎𝑐𝑘𝑒𝑡 ⇒ 4 𝐶ℎ𝑜𝑖𝑐𝑒𝑠

𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑓𝑜𝑟 𝑇𝑖𝑒: 𝑃𝑜𝑙𝑘𝑎 𝐷𝑜𝑡𝑡𝑒𝑑, 𝑆𝑡𝑟𝑖𝑝𝑒𝑑, 𝑁𝑜 𝑇𝑖𝑒 ⇒ 3 𝐶ℎ𝑜𝑖𝑐𝑒𝑠

4 × 3 = 12
E. Routes between Cities

(Important) Example 4.13

Mumbai is connected to Alibaugh via road, ferry, and train. A person will travel using exactly one of these
methods for a journey from one city to another. The return journey can be via a different method. In how many
ways can a person:
A. Go to Alibaugh from Mumbai
B. Go to Mumbai from Alibaugh
C. Go to Alibaugh from Mumbai and come back given that:
a. There are no restrictions
b. You come back via the same route that you went.
c. You come back via a different way from the one that you went by.

Part A and B
⏟
1 + ⏟
1 + ⏟
1 =3 Road Ferry Train
𝑅𝑜𝑎𝑑 𝐹𝑒𝑟𝑟𝑦 𝑇𝑟𝑎𝑖𝑛 Road
Part C Ferry
If there are no restrictions: Train
𝑎: ⏟
3 × ⏟
3 =9
𝑮𝒐𝒊𝒏𝒈 𝑪𝒐𝒎𝒊𝒏𝒈
𝑩𝒂𝒄𝒌
If you come back via the same route that you went.
𝑏: ⏟3 × ⏟
1 =3
𝑮𝒐𝒊𝒏𝒈 𝑪𝒐𝒎𝒊𝒏𝒈
𝑩𝒂𝒄𝒌
If you come back via a different way from the one that you went by.
𝑐: ⏟𝟑 × ⏟ 2 =6
𝑮𝒐𝒊𝒏𝒈 𝑪𝒐𝒎𝒊𝒏𝒈
𝑩𝒂𝒄𝒌

Example 4.14
Answer the same questions as above, except that now there are three roads, two ferries and five trains.
(Assume that the roads, ferries, and trains are all distinct, and if you come back a different way from the one
that you went by, you can still repeat the same type of travel. For example, you can go by Road 1,but come back
by Road 2.)

Parts A and B:
⏟
3 + ⏟
2 + ⏟
5 = 10
𝑅𝑜𝑎𝑑 𝐹𝑒𝑟𝑟𝑦 𝑇𝑟𝑎𝑖𝑛
Part C:
𝐼: ⏟
3+2+5× ⏟
3 + 2 + 5 = 10 × 10 = 100
𝐺𝑜𝑖𝑛𝑔 𝐶𝑜𝑚𝑖𝑛𝑔 𝐵𝑎𝑐𝑘
𝐼𝐼: ⏟
3+2+5× ⏟
1 = 10 × 1 = 10
𝐺𝑜𝑖𝑛𝑔 𝐶𝑜𝑚𝑖𝑛𝑔 𝐵𝑎𝑐𝑘

If you can repeat the type of travel, then you get:

𝐼𝐼𝐼: ⏟
3+2+5×⏟ 3 + 2 + 5 − 1 = 10 × 9 = 90
𝐺𝑜𝑖𝑛𝑔 𝐶𝑜𝑚𝑖𝑛𝑔 𝐵𝑎𝑐𝑘

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Example 4.15
There are three highways that go from Denver to Philadelphia, and six flights that go from Philadelphia to
Washington and 𝑣𝑖𝑐𝑒 𝑣𝑒𝑟𝑠𝑎. Due to a snowstorm, the only route between Denver to Washington is via
Philadelphia.
A. What is the number of ways to go from Denver to Washington?
B. What is the number of ways to go from Denver to Washington and then come back if:
a. You come back the same way that you went
b. You come back via any way that you want
c. You come back, but not via the exact same way that you took to reach Washington. (If you went
by Highway 1, followed by Flight 2, coming back via anything except Flight 2, followed by
Highway 1 is acceptable.)

Part A:
⏟
3 × ⏟
6 ⇒ 3 × 6 = 18
𝑾𝒂𝒚𝒔 𝒕𝒐 𝒈𝒐 𝒇𝒓𝒐𝒎 𝑾𝒂𝒚𝒔 𝒕𝒐 𝒈𝒐 𝒇𝒓𝒐𝒎
𝑫𝒆𝒏𝒗𝒆𝒓 𝒕𝒐 𝑷𝒉𝒊𝒍𝒂𝒅𝒆𝒍𝒑𝒉𝒊𝒂 𝑷𝒉𝒊𝒍𝒂𝒅𝒆𝒍𝒑𝒉𝒊𝒂 𝒕𝒐 𝑾𝒂𝒔𝒉𝒊𝒏𝒈𝒕𝒐𝒏𝒂

Part B
Sub-Part a
18
⏟ × ⏟
1 = 18
𝑾𝒂𝒚𝒔 𝒕𝒐 𝒈𝒐 𝒇𝒓𝒐𝒎 𝑾𝒂𝒚𝒔 𝒕𝒐 𝒈𝒐 𝒇𝒓𝒐𝒎
𝑫𝒆𝒏𝒗𝒆𝒓 𝒕𝒐 𝑾𝒂𝒔𝒉𝒊𝒏𝒈𝒕𝒐𝒏 𝑾𝒂𝒔𝒉𝒊𝒏𝒈𝒕𝒐𝒏 𝒕𝒐 𝑫𝒆𝒏𝒗𝒆𝒓
Sub-Part b
18
⏟ × 18
⏟ ⇒ 18 × 18 = 182 = 324
𝑾𝒂𝒚𝒔 𝒕𝒐 𝒈𝒐 𝒇𝒓𝒐𝒎 𝑾𝒂𝒚𝒔 𝒕𝒐 𝒈𝒐 𝒇𝒓𝒐𝒎
𝑫𝒆𝒏𝒗𝒆𝒓 𝒕𝒐 𝑾𝒂𝒔𝒉𝒊𝒏𝒈𝒕𝒐𝒏 𝑾𝒂𝒔𝒉𝒊𝒏𝒈𝒕𝒐𝒏 𝒕𝒐 𝑫𝒆𝒏𝒗𝒆𝒓
Sub-Part c
18
⏟ × 17
⏟ = 306
𝑾𝒂𝒚𝒔 𝒕𝒐 𝒈𝒐 𝒇𝒓𝒐𝒎 𝑾𝒂𝒚𝒔 𝒕𝒐 𝒈𝒐 𝒇𝒓𝒐𝒎
𝑫𝒆𝒏𝒗𝒆𝒓 𝒕𝒐 𝑾𝒂𝒔𝒉𝒊𝒏𝒈𝒕𝒐𝒏 𝑾𝒂𝒔𝒉𝒊𝒏𝒈𝒕𝒐𝒏 𝒕𝒐 𝑫𝒆𝒏𝒗𝒆𝒓

F. Deciding between addition and multiplication

Example 4.16: Travelling by road or by flight

Han wants to reach Toronto from Ontario. He can take any one of five roads to reach Toronto. Or, he can pick
one of three flights to reach Toronto. What is the number of ways in which he can reach Toronto?

𝑅𝑜𝑎𝑑𝑠
⏟ + 𝐹𝑙𝑖𝑔ℎ𝑡𝑠
⏟ ⇒5+3=8
𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 𝟑 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Example 4.17: Exam Scenario

𝐴𝑛𝑠𝑤𝑒𝑟 𝑒𝑎𝑐ℎ 𝑝𝑎𝑟𝑡 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑒𝑙𝑦
An exam with seven subjective (free response) and three objective (multiple choice) questions can be
attempted in one of four different ways. The details are, by answering exactly:
A. one subjective question
B. one objective question
C. one subjective and one objective question
D. one subjective or one objective question

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What is the number of ways in which the paper can be attempted under the different options?

Option A
There are seven subjective questions, of which exactly one must be attempted. Hence, I have
7 𝑄𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠 = 7 𝑊𝑎𝑦𝑠
Option B
There are three objective questions, of which exactly one must be attempted. Hence, I have
3 𝑄𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠 = 3 𝑊𝑎𝑦𝑠
Option C
We must combine one subjective question with one objective question. By the multiplication principle, this can
be done in
⏟
7 × ⏟
3 = 21 𝑊𝑎𝑦𝑠
𝑆𝑢𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑂𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒
𝑄𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠 𝑄𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠
Option D
We must answer either a subjective question or an objective question. By the addition principle, this can be
done in
⏟
7 + ⏟
3 = 10 𝑊𝑎𝑦𝑠
𝑆𝑢𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑂𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒
𝑄𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠 𝑄𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠

4.2 Using the Rules

A. Extending the Multiplication Principle
The multiplication principle can be extended from two making a choice that has two parts to many parts. In
addition, it is important to recognize when not doing something is also a choice.

4.18: Extending the Multiplication Principle

If I can do event A in 𝑎 ways, event B in 𝑏 ways, and event C in 𝑐 ways, and event D in 𝑑 ways, then I can do all of
them together in
𝑎𝑏𝑐𝑑 𝑤𝑎𝑦𝑠

Example 4.19
A. I am going to a party, and I need to choose my shoes, socks, jacket, and my tie. I have an ochre jacket, a
maroon jacket, and a green jacket. I have a polka dotted tie, and a striped tie. I have shoes in three
colours: black, brown, and white. I have socks in four colours: gray, blue, yellow and red. In how many
ways can I choose my outfit if I always wear shoes but have the option of not wearing one or more of my
jackets, my ties, and my socks.
B. I have two jackets, four shirts, four pairs of trousers, two pairs of shoes, and only one tie. How many
outfits can I make consisting of a shirt, a pair of trousers, a pair of shoes, an (optional) tie, and an
(optional) jacket?
C. Items at a restaurant are available in any of three bases (unless specified otherwise): 𝑉𝑒𝑔𝑒𝑡𝑎𝑟𝑖𝑎𝑛,
𝐶ℎ𝑖𝑐𝑘𝑒𝑛 and 𝑆𝑒𝑎𝑓𝑜𝑜𝑑. In main course, I can order 𝑆𝑖𝑐ℎ𝑢𝑎𝑛 or 𝑀𝑎𝑛𝑐ℎ𝑢𝑟𝑖𝑎𝑛, while for soup, I can
choose between four options: 𝐻𝑜𝑡 & 𝑆𝑜𝑢𝑟, 𝑊𝑖𝑛𝑡𝑒𝑟 𝑀𝑒𝑙𝑜𝑛, 𝑊𝑜𝑛𝑡𝑜𝑛 and 𝑆𝑙𝑜𝑤 𝐶𝑜𝑜𝑘𝑒𝑑. In how many
ways can I order a soup and a main course, given that I can (optionally) order dry noodles as a side dish
(dry noodles are available in only one variety – no choice of base)

Part A
⏟
3 × ⏟
5 × ⏟
4 × ⏟
3 = 180
𝑺𝒉𝒐𝒆𝒔 𝑺𝒐𝒄𝒌𝒔 𝑱𝒂𝒄𝒌𝒆𝒕𝒔 𝑻𝒊𝒆𝒔

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Part B
⏟
3 × ⏟
4 × ⏟
4 × ⏟
2 × ⏟
2 = 192
𝑱𝒂𝒄𝒌𝒆𝒕 𝑺𝒉𝒊𝒓𝒕 𝑻𝒓𝒐𝒖𝒔𝒆𝒓𝒔 𝑺𝒉𝒐𝒆𝒔 𝑻𝒊𝒆𝒔
Part C
Soup
⏟ × 𝑀𝑎𝑖𝑛 𝐶𝑜𝑢𝑟𝑠𝑒
⏟ × 𝑆𝑖𝑑𝑒 𝐷𝑖𝑠ℎ
⏟ = 12 × 6 × 2 = 144
𝟒 𝑽𝒂𝒓𝒊𝒆𝒕𝒊𝒆𝒔×𝟑 𝑩𝒂𝒔𝒆𝒔=𝟏𝟐 𝟐 𝑽𝒂𝒓𝒊𝒆𝒕𝒊𝒆𝒔×𝟑 𝑩𝒂𝒔𝒆𝒔=𝟔 𝟐 𝑪𝒉𝒐𝒊𝒄𝒆𝒔
𝑶𝒓𝒅𝒆𝒓 𝒐𝒓 𝒅𝒐𝒏′ 𝒕 𝒐𝒓𝒅𝒆𝒓

Example 4.20: Double Decker Sandwich

A sandwich has three layers of bread with two fillings (Bread / Filling / Bread Veg Non-veg Butter
/ Filling / Bread). The types of fillings and types of butter are mentioned in Aloo Chicken Plain
the table. The location of the filling matters, and bread in touch with a filling is Salad Fish Garlic
buttered. Peas
A. For each sandwich, how many sides will I butter?
B. What is the number of distinct sandwiches that can be made if one filling must be veg, and the other
filling must be non-veg, and butter in a single sandwich is the same for all sides? (Note: Two sandwiches
which can be flipped over to make them the same are still considered distinct).
C. How many sides will I butter to make each distinct sandwich once if:
a) Fillings can be repeated, and butter in a single sandwich is the same for all sides.
b) Fillings cannot be repeated, but butter in a single sandwich does not have to be the same for all
sides, but any one side uses a single type of butter.

Part A
Three layers of bread will have Top Layer Top Side
6 𝑆𝑖𝑑𝑒𝑠 Bottom Side
Out of the 6 sides, all except the topmost and the bottommost will be Filling
buttered. Hence, the number of sides to be buttered Middle Layer Top Side
= 𝑇𝑜𝑡𝑎𝑙 𝑆𝑖𝑑𝑒𝑠 − 𝑆𝑖𝑑𝑒𝑠 𝑁𝑜𝑡 𝐵𝑢𝑡𝑡𝑒𝑟𝑒𝑑 = 6 − 2 = 4 Bottom Side
Part B Filling
Number of Distinct Sandwiches: Bottom Layer Top Side
Bottom Side
⏟
3 × ⏟
2 × ⏟ 2 = 12
𝑽𝒆𝒈 𝑵𝒐𝒏−𝑽𝒆𝒈 𝑩𝒖𝒕𝒕𝒆𝒓
𝑭𝒊𝒍𝒍𝒊𝒏𝒈 𝑭𝒊𝒍𝒍𝒊𝒏𝒈
Location of Filling:
We also have a choice of deciding whether the topmost filling is veg, or the bottommost filling is veg, giving us
two choices:

Bread 𝐶ℎ𝑜𝑖𝑐𝑒 𝐼 𝐶ℎ𝑜𝑖𝑐𝑒 𝐼𝐼

Filling Veg Non-Veg
Bread
Filling Non-Veg Veg
Bread

Hence the final answer is:

12 × 2 = 24
Part C-I: Fillings can be repeated
Number of Distinct Sandwiches
⏟
5 × ⏟
5 × ⏟
2 = 50
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑩𝒖𝒕𝒕𝒆𝒓
𝑭𝒊𝒍𝒍𝒊𝒏𝒈 𝑭𝒊𝒍𝒍𝒊𝒏𝒈

Total Sides Buttered:

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𝑇𝑜𝑡𝑎𝑙 𝑆𝑖𝑑𝑒𝑠 𝐵𝑢𝑡𝑡𝑒𝑟𝑒𝑑 = 50 × 4 = 200

Part C-II: Fillings cannot be repeated

Count the number of options when buttering the sides:
⏟
2 × ⏟ 2 × ⏟ 2 × ⏟
2 = 16
𝑺𝒊𝒅𝒆 𝟏 𝑺𝒊𝒅𝒆 𝟏 𝑺𝒊𝒅𝒆 𝟏 𝑺𝒊𝒅𝒆 𝟏
Number of Distinct Sandwiches
⏟
5 × ⏟
4 × 16
⏟ = 320
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑩𝒖𝒕𝒕𝒆𝒓
𝑭𝒊𝒍𝒍𝒊𝒏𝒈 𝑭𝒊𝒍𝒍𝒊𝒏𝒈
Total Sides Buttered
𝑇𝑜𝑡𝑎𝑙 𝑆𝑖𝑑𝑒𝑠 𝐵𝑢𝑡𝑡𝑒𝑟𝑒𝑑 = 320 × 4 = 1,280

Example 4.21: Filling Posts

A MUN committee has 10 representatives from Asia, 8 from Europe, and 7 from Africa. If a representative is
eligible for a maximum of one post, in how many ways can a 𝑆𝑒𝑐𝑟𝑒𝑡𝑎𝑟𝑦
⏟ 𝐺𝑒𝑛𝑒𝑟𝑎𝑙, a 𝐷𝑒𝑝𝑢𝑡𝑦
⏟ 𝑆𝑒𝑐𝑟𝑒𝑡𝑎𝑟𝑦 𝐺𝑒𝑛𝑒𝑟𝑎𝑙
𝑺𝑮 𝐷𝑆𝐺
and a 𝑃𝑟𝑒𝑠𝑖𝑑𝑒𝑛𝑡
⏟ be chosen for the committee if:
𝑃
A. There are no further restrictions
B. The Secretary General must be from Asia, the Deputy Secretary General must from Europe and the
President must be from Africa
C. They must each be from different continents
D. They must not all be from the same continent
E. Exactly two of them must be from the same continent

Total Number of Representatives

10
⏟ + ⏟
8 + ⏟
7 = 25
𝐴𝑠𝑖𝑎 𝐸𝑢𝑟𝑜𝑝𝑒 𝐴𝑓𝑟𝑖𝑐𝑎

Part A
25
⏟ × 24
⏟ × 23
⏟ = 13,800
𝑆𝐺 𝐷𝑆𝐺 𝑃

Part B
10
⏟× ⏟
8 ×⏟
7 = 560
𝑆𝐺 𝐷𝑆𝐺 𝑃

Part C
We first need to allocate the continent to the post. This can be done in:
3 𝐶𝑜𝑛𝑡𝑖𝑛𝑒𝑛𝑡𝑠 × ⏟
⏟ 2 𝑐𝑜𝑛𝑡𝑖𝑛𝑒𝑛𝑡𝑠 × ⏟1 𝑐𝑜𝑛𝑡𝑖𝑛𝑒𝑛𝑡 = 6
𝑆𝐺 𝐷𝑆𝐺 𝑃
Then, whichever continent we have allocated to the post, we need to select the representative who will fill that
post:
10
⏟ × ⏟ 8 × ⏟ 7 = 560
𝐴𝑠𝑖𝑎 𝐸𝑢𝑟𝑜𝑝𝑒 𝐴𝑓𝑟𝑖𝑐𝑎
Finally, the two choices above can be combined to get:
560 × 6 = 3,360
Part D
Counting all the different cases where they are not all from the same continent will be a lot of casework. Instead,
we use complementary counting.

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All posts from the same continent

𝐴𝑙𝑙 𝑓𝑟𝑜𝑚 𝐴𝑠𝑖𝑎: 10 × 9 × 8 = 720
𝐴𝑙𝑙 𝑓𝑟𝑜𝑚 𝐸𝑢𝑟𝑜𝑝𝑒: 8 × 7 × 6 = 336
𝐴𝑙𝑙 𝑓𝑟𝑜𝑚 𝐴𝑓𝑟𝑖𝑐𝑎: 7 × 6 × 5 = 210
𝑇𝑜𝑡𝑎𝑙: 720 + 336 + 210 = 1,266

13,800
⏟ − 1,266
⏟ = 12,534
𝑇𝑜𝑡𝑎𝑙 𝑊𝑎𝑦𝑠 𝐴𝑙𝑙 𝑓𝑟𝑜𝑚 𝑠𝑎𝑚𝑒
𝐶𝑜𝑛𝑡𝑖𝑛𝑒𝑛𝑡

Part E: Exactly two must be from the same continent

13,800
⏟ − 1,266
⏟ − ⏟
3360 = 9174
𝑇𝑜𝑡𝑎𝑙 𝑊𝑎𝑦𝑠 𝐴𝑙𝑙 𝑓𝑟𝑜𝑚 𝑠𝑎𝑚𝑒 𝐴𝑙𝑙 𝑡ℎ𝑟𝑒𝑒
𝐶𝑜𝑛𝑡𝑖𝑛𝑒𝑛𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑐𝑜𝑛𝑡𝑖𝑛𝑒𝑛𝑡𝑠

3 + 2𝑥 < 𝑥 𝐴𝑁𝐷 2 + 𝑥 > 8

B. Back Calculations

Example 4.22
A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an
appetizer, a main course, and a dessert. What is the least number of main courses that a restaurant should offer
so that a customer could have a different dinner each night in the year 2003? (AMC 10B 2003/16)

2003 is a non-leap year. Hence, the number of Days in 2003 is

365
Let the number of main courses be 𝑥. Then, the number of different dinners is:
= ⏟
3 × 2𝑥
⏟ × ⏟ 𝑥 = 6𝑥 2
𝑫𝒆𝒔𝒔𝒆𝒓𝒕𝒔 𝑨𝒑𝒑𝒆𝒕𝒊𝒛𝒆𝒓𝒔 𝑴𝒂𝒊𝒏
𝑪𝒐𝒖𝒓𝒔𝒆𝒔
But the number of dinners has to be greater than or equal to 365.

Method I: Solve using Guess and Check

If you don’t know inequalities, the values are small enough for you to get via trial and error. 7 is too small, and
hence 8 is the smallest number that satisfies the condition.
6 × 72 = 294 < 365, 6 × 82 = 384 > 365

Method II: Solve using Inequalities

365 5 5
6𝑥 2 ≥ 365 ⇒ 𝑥 2 ≥ = 60 ⇒ 72 = 49 < 60 < 64 = 82 ⇒ 𝑆𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑥 𝑖𝑠 8
6 6 6
C. Creating Positions
Many times, when we want to count the number of ways, something can be arranged, we look at the positions.
Sometimes, it may be useful to create positions to make our counting easy.
The following examples show this.

Example 4.23: Wearing Two Rings

I am going to a party wearing a total of two rings on the five appendages of my right hand (four fingers, and one
thumb, with no restrictions). I can wear more than one ring on an appendage. Also, the order of the rings
matters. In how many ways can I wear two rings?

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Method I: Casework
Case I: Two Rings on the same finger
⏟
𝟓 × ⏟
𝟐 = 𝟏𝟎
𝑵𝒐.𝒐𝒇 𝑨𝒓𝒓𝒂𝒏𝒈𝒊𝒏𝒈
𝑨𝒑𝒑𝒆𝒏𝒅𝒂𝒈𝒆𝒔 𝑻𝒘𝒐 𝑹𝒊𝒏𝒈𝒔
Case II: Two Rings on two different fingers
⏟
5 × ⏟
4 = 20
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅
𝑨𝒑𝒑𝒆𝒏𝒅𝒂𝒈𝒆 𝑨𝒑𝒑𝒆𝒏𝒅𝒂𝒈𝒆

𝑇𝑜𝑡𝑎𝑙 𝑊𝑎𝑦𝑠 = 10 + 20 = 30
Method II: Multiplication Principle
Wear the first ring. This can be done on any of the fingers, or the thumb, giving us
5 𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑠
Now, the second ring can go on any finger or thumb. On whichever finger the first ring was worn, it can go in
two positions. So, the total number of positions for the second ring is:
6 𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑠
By the multiplication principle, the total number of ways of wearing both rings is:
⏟
5 × ⏟ 6 = 30
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅
𝑹𝒊𝒏𝒈 𝑹𝒊𝒏𝒈

Example 4.24: Wearing Three Rings

For the next party, I wear three rings instead of two, order still matters, and there are no restrictions on how I
can wear the rings. In how many ways can I do this?

Multiplication Principle
⏟
5 × ⏟
6 × ⏟
7 = 210
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑻𝒉𝒊𝒓𝒅
𝑹𝒊𝒏𝒈 𝑹𝒊𝒏𝒈 𝑹𝒊𝒏𝒈
Casework
Case I: All Three Rings on one Finger
Case II: Two Rings on One Finger, and One Ring on a Different Finger
Case III: Three Rings on Three Different Fingers

D. Pairs

4.25: Pair
If we put two things together, they form a pair.

For example:
(5,7) ⇒ 𝑆𝑢𝑐𝑐𝑒𝑠𝑠𝑖𝑣𝑒 𝑂𝑑𝑑 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
(𝑥, 𝑦) ⇒ 𝐶𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑖𝑛 𝑎 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
(𝐷𝑒𝑙ℎ𝑖, 𝐼𝑛𝑑𝑖𝑎) ⇒ 𝐶𝑖𝑡𝑦 𝑎𝑛𝑑 𝑖𝑡𝑠 𝑐𝑜𝑢𝑛𝑡𝑟𝑦
(𝑆𝑎𝑐ℎ𝑖𝑛 𝑇𝑒𝑛𝑑𝑢𝑙𝑘𝑎𝑟, 𝐶𝑟𝑖𝑐𝑘𝑒𝑡) ⇒ 𝑆𝑝𝑜𝑟𝑡𝑠𝑝𝑒𝑟𝑠𝑜𝑛 𝑎𝑛𝑑 𝑆𝑝𝑜𝑟𝑡

4.26: Unordered Pair

If the sequence of the pair is not important, then the pair is called an unordered pair. In such a case
(5,7) = (7,5)

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4.27: Ordered Pair

On the other hand, if sequence is important, then the pair is an ordered pair. Then
(5,7) ≠ (7,5)

E. Number of Subsets

Example 4.28
List the subsets of the set {𝐴, 𝐵, 𝐶} including the set itself, and the set with no elements (null set).
Recall that the subset of a set 𝑋 is a set that does not have any elements which are not present in 𝑋.

Consider only C
We will build up the solution step by step. Consider that we had a set with a single element 𝐶. We have a choice
of taking that element, or not taking it.

C {C}
~C {}
Consider B and C
With each of B and C, I have a choice of picking or not picking the element. I can combine the choices into a tree
diagram.

C {B,C}
B
~C {B}
C {C}
~B
~C {}
Consider A, B and C
IF we consider all three elements, with each of these, I have a choice of picking or not picking the element. I can
represent these using a much larger tree diagram, which tells me that I have 8 choices in all.

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C {A,B,C}
B
~C {A,B}
A
C {A, C}
~B
~C {A}
C {B,C}
B
~C {B}
~A
C {C}
~B
~C {}

{{𝐴, 𝐵, 𝐶}, {𝐴, 𝐵}, {𝐴, 𝐶}, {𝐴}, {𝐵, 𝐶}, {𝐵}, {𝐶}, {𝜙} }

Find the number of different subsets listed above:

A. By counting
B. By using the multiplication rule

Counting
𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 8 𝑠𝑢𝑏𝑠𝑒𝑡𝑠

My choices are:
⏟
2 × ⏟
2 × ⏟
2 = 23 = 8
𝑇𝑤𝑜 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑓𝑜𝑟 𝐴: 𝑇𝑤𝑜 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑓𝑜𝑟 𝐵: 𝑇𝑤𝑜 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑓𝑜𝑟 𝐶:
𝑇𝑎𝑘𝑒 𝑜𝑟 𝐷𝑜𝑛′ 𝑡 𝑇𝑎𝑘𝑒 𝑇𝑎𝑘𝑒 𝑜𝑟 𝐷𝑜𝑛′ 𝑡 𝑇𝑎𝑘𝑒 𝑇𝑎𝑘𝑒 𝑜𝑟 𝐷𝑜𝑛′ 𝑡 𝑇𝑎𝑘𝑒

4.29: Number of Subsets

The number of subsets of a set with 𝑛 elements is:
2𝑛

Example 4.30
What is the number of subsets of a set which contains:
A. 20 elements?
B. as its elements the capital letters of the English alphabet?

220
226

Example 4.31
7 prizes, named 𝐴, 𝐵, … , 𝐺 have to be distributed among 2 boys based on their performance in a sport

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competition. What is the number of ways in which the prizes can be distributed?

The prizes are

{𝐴, 𝐵, 𝐶, 𝐷, 𝐸, 𝐹, 𝐺}
Consider making subsets of the prizes, and let the first boy get the prizes in that subset.
Then, the number of ways of distributing the prizes is the same as the number of subsets of a set with 7
elements:
= 27 = 128

Alternate Solution
Prize 𝐴 can be given either to the first boy, or the second boy. This gives us
2 𝑐ℎ𝑜𝑖𝑐𝑒𝑠
Similarly, Prize 𝐵 can be given either to the first boy, or the second boy. This also gives us:
2 𝑐ℎ𝑜𝑖𝑐𝑒𝑠
We have two choices for each prize, giving us:
⏟
2 × ⏟ 2 × …× ⏟ 2 = 27 = 128 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑 𝑆𝑒𝑣𝑒𝑛𝑡ℎ
𝑃𝑟𝑖𝑧𝑒 𝑃𝑟𝑖𝑧𝑒 𝑃𝑟𝑖𝑧𝑒

Example 4.32
Two leading scientists have been nominated for 9 awards, numbered 1, 2, … ,9. In order to not upset the
competitive scientists, each scientist must receive at least one award. If they are the only ones who have been
nominated for these awards, then in how many ways can the awards be distributed?

Consider making subsets of the awards, and let the first scientist get the awards in that subset.
Then, the number of ways of distributing the awards is the same as the number of subsets of a set with 9
elements:
= 29 = 512

But we cannot have a null set (no prizes to Scientist A), or the entire set (no prizes to Scientist B).
Hence, the final answer is
512 − 2 = 510

Alternate Solution

We use the concept of complementary counting. Imagine there are no restrictions. Then, as in the previous
example, the number of ways to distribute the awards is:
⏟
2 × ⏟ 2 × …× ⏟ 2 = 29 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑 𝑁𝑖𝑛𝑡ℎ
𝐴𝑤𝑎𝑟𝑑 𝐴𝑤𝑎𝑟𝑑 𝐴𝑤𝑎𝑟𝑑
But, we have to give at least one award to each scientist. Hence, there are two ways in which we cannot
distribute the awards
𝑊𝑎𝑦 1: 𝐴𝑙𝑙 𝑎𝑤𝑎𝑟𝑑𝑠 𝑡𝑜 𝑓𝑖𝑟𝑠𝑡 𝑠𝑐𝑖𝑒𝑛𝑡𝑖𝑠𝑡
𝑊𝑎𝑦 2: 𝐴𝑙𝑙 𝑎𝑤𝑎𝑟𝑑𝑠 𝑡𝑜 𝑠𝑒𝑐𝑜𝑛𝑑 𝑠𝑐𝑖𝑒𝑛𝑡𝑖𝑠𝑡

Hence, the final answer is:

29 − 2 = 512 − 2 = 510

Example 4.33
Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the
guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of

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guides and tourists are possible? (AMC 10A 2004/12)

You can think about this question in the context of the powerset formulas that we just built up.
Consider that the guides are:
𝐴 𝑎𝑛𝑑 𝐵

And that you are considering the set of tourists that will go with A.
Then, the total number of choices you can make is the same as the number of non-null, proper subsets that you
can make from the set of six tourists
26 − 1 − 1 = 64 − 2 = 62

Alternate Solution
Consider the number of choices for each tourist:
⏟2 × ⏟
2 × …× ⏟
2 = 26 = 64
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑 𝑆𝑖𝑥𝑡ℎ
𝑇𝑜𝑢𝑟𝑖𝑠𝑡: 𝑇𝑜𝑢𝑟𝑖𝑠𝑡: 𝑇𝑜𝑢𝑟𝑖𝑠𝑡:
𝑇𝑤𝑜 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑇𝑤𝑜 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑇𝑤𝑜 𝐶ℎ𝑜𝑖𝑐𝑒𝑠

64 − 2 = 62
A. Powerset

4.34: Powerset
The powerset consists of the set of all subsets of a set.

Example 4.35
Consider the set
{𝐴, 𝐵, 𝐶}
A. What is the powerset of the set
B. What is the cardinality of the powerset?

We already did this, without using the term powerset:

{{𝐴, 𝐵, 𝐶}, {𝐴, 𝐵}, {𝐴, 𝐶}, {𝐴}, {𝐵, 𝐶}, {𝐵}, {𝐶}, {𝜙}}

And, we also calculated the cardinality of the above set, which is:
23 = 8

4.36: Cardinality of Powerset

Cardinality is the number of elements of a set.

The cardinality of the powerset of a set 𝐴 with 𝑛 elements is given by:

2𝑛

⏟
2 × ⏟
2 × …× ⏟
2 = 2𝑛
2 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 2 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 2 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
𝑓𝑜𝑟 𝑡ℎ𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒
1𝑠𝑡 𝐸𝑙𝑒𝑚𝑒𝑛𝑡 2𝑛𝑑 𝐸𝑙𝑒𝑚𝑒𝑛𝑡 𝑛𝑡ℎ 𝐸𝑙𝑒𝑚𝑒𝑛𝑡

4.37: Null and Non-Null Sets

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An empty set is called a null set.

A non-empty set is called a non-null set.

𝐸𝑚𝑝𝑡𝑦 𝑆𝑒𝑡 = {𝜙}

4.38: Proper Subset

A subset of a set, which is not the set itself is called a proper subset.

Example 4.39
A. How subsets of a set are not proper?

𝐸𝑥𝑎𝑐𝑡𝑙𝑦 𝑂𝑛𝑒

Example 4.40
Mary has five treasured family heirlooms to be distributed among her two children, A and B. She is fine with any
distribution that does not require all of the heirlooms to be given to a single child. In how many ways can she
make this distribution?

Let the heirlooms be:

𝐻 = {ℎ1 , ℎ2 , ℎ3 , ℎ4 , ℎ5 }

We can make a division of the heirlooms in various ways. For example, if we decide to give heirlooms ℎ2 and ℎ5
to A, then we will get the distribution:
𝐴 = {ℎ2 , ℎ5 }, 𝐵 = {ℎ1 , ℎ3 , ℎ4 }

In other words, giving heirlooms to A is the same as making a subset of H.

The number of ways we can make subsets of H is:
25

However, we cannot have the following distributions:

𝐴 = {𝜙}, 𝐵 = {ℎ1 , ℎ2 , ℎ3 , ℎ4 , ℎ5 }
𝐴 = {ℎ1 , ℎ2 , ℎ3 , ℎ4 , ℎ5 }, 𝐵 = {𝜙}

The final answer is:

25 − 2 = 32 − 2 = 30

Example 4.41
Consider the set that has the letters of the English Alphabet:
𝐿 = {𝐴, 𝐵, 𝐶, … , 𝑍}
A. What is the cardinality of the powerset?
B. What is the number of non-null elements of the powerset?
C. What is the number of proper subsets of L?
D. What is the number of non-null, proper subsets of L?

226
26
2 −1
226 − 1
2 − 1 − 1 = 226 − 2
26

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Repeat the above question with the set of digits in the decimal system:
𝐷 = {0,1,2, … ,9}

210
210 − 1
210 − 1
210 − 1 − 1 = 226 − 2
4.3 Repetition
A. Repetition Not Allowed
The questions seen so far required picking an element from Set 𝐴, an element from Set 𝐵, and so on. We now
look at questions where Set 𝐴 and Set 𝐵 are actually the same set.

4.42: Repetition Not Allowed

If the selection is from the same set, the question arises, whether you can select the same element twice. The
choice of whether repetition is allowed may be
➢ Explicit: Made clear in the question itself.
➢ Implicit: Because the question gives a real-life scenario where repetition would not be meaningful

The examples below show questions where repetition is not allowed because the question specifically forbids it.

Example 4.43
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window
and leave by a different window? (AMC 8 2007/4)

The choices for entering and the choice for leaving are both from the same set. But repetition is not allowed.
Hence, the number of choices for leaving will be one less than the number of choices for entering:
𝐸𝑛𝑡𝑒𝑟𝑖𝑛𝑔
⏟ × 𝐿𝑒𝑎𝑣𝑖𝑛𝑔
⏟ = 6 × 5 = 30
𝟔 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 𝟔−𝟏= 𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Entering No. of
1 2 3 4 5 6 Choices
Leaving 1 5
2 5
3 5
4 5
5 5
6 5

5 × 6 = 30

Example 4.44: Entering and Leaving a City

Victor is a spy entering the city of Transylvania. He has the option of entering by road, by sea, or by flight. He
knows that the police will get alerted once he enters a city via a route, so he cannot leave by the same route that
he enters. How many different routes can he plan to enter and exit the city?

Since the choice

(𝑅𝑜𝑎𝑑, 𝑆𝑒𝑎) ≠ (𝑆𝑒𝑎, 𝑅𝑜𝑎𝑑)
we need the number of ordered pairs.

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Victor has three choices when he enters. When he leaves, he has one less choice. The choice of route that he
chooses to enter does not affect this.
{𝑅𝑜𝑎𝑑, 𝑆𝑒𝑎, 𝐹𝑙𝑖𝑔ℎ𝑡} ×
⏟ {⏟} ⇒3×2=6
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑾𝒂𝒚𝒔 𝒐𝒇 𝑬𝒏𝒕𝒆𝒓𝒊𝒏𝒈 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒘𝒂𝒚𝒔 𝒐𝒇 𝒍𝒆𝒂𝒗𝒊𝒏𝒈
=𝟑 =𝟐

Example 4.45: Routes between Cities

How many ways are there to go from City 𝐴 to City 𝐵 (three routes), then City 𝐵 to City 𝐶 (four routes) and then
come back to City A via City 𝐵, but via a different way from the one used to travel to City 𝐶? (Two routes are the
same, if and only if they use the same route from 𝐴 to 𝐵, and also from 𝐵 to 𝐶, and the other way around).

Going from City A to City C:

City A to City B City B to City C

𝑋1 𝑌1 {𝑋1 𝑌1 , 𝑋1 𝑌2 , 𝑋1 𝑌3 , 𝑋1 𝑌4 } → 4 𝑅𝑜𝑢𝑡𝑒𝑠
𝑋2 𝑌2 {𝑋2 𝑌1 , 𝑋2 𝑌2 , 𝑋2 𝑌3 , 𝑋2 𝑌4 } → 4 𝑅𝑜𝑢𝑡𝑒𝑠
𝑋3 𝑌3 {𝑋3 𝑌1 , 𝑋3 𝑌2 , 𝑋3 𝑌3 , 𝑋3 𝑌4 } → 4 𝑅𝑜𝑢𝑡𝑒𝑠
𝑌4

Total Ways 3 4 3 × 4 = 12

𝐴⏟→ 𝐵 × 𝐵
⏟→ 𝐶 ⇒ 𝐴 → 𝐶 = 3 × 4 = 12 𝑅𝑜𝑢𝑡𝑒𝑠
3 4
Going from City C to City A:
𝐴 → 𝐶 = 3 × 4 = 12
But we cannot go back via the same route that we came from A to C. Hence, we have one route less
12 − 1 = 11

𝐴
⏟→ 𝐶 × 𝐶⏟→ 𝐴 = 12 × 11 = 132
12 11

B. Repetition Understood from the Context

In real-life scenarios, the situation may tell us whether repetition is allowed. In such cases, the question may not
tell this to us directly.

Selecting two elements from a set 𝐴 with 𝑛 elements gives:

(𝑎, 𝑏)

Example 4.46: Selecting Monitors

A class has 12 boys and 7 girls. Find the number of ways to choose:
A. A monitor.
B. A monitor of each gender.
C. A monitor and a co-monitor (where a student cannot be both a monitor and a co-monitor at the same
time).
D. A monitor and a co-monitor, where one of them is a boy, and the other is a girl.

Part A
There are no restrictions on the gender of the monitor. Hence, the monitor can be any of the students of the
class, which is:
12
⏟ + ⏟ 7 = 19 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 ⇒ 19 𝑊𝑎𝑦𝑠
𝑩𝒐𝒚𝒔 𝑮𝒊𝒓𝒍𝒔

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Part B
One monitor must be a boy (12 Choices), and the other must be a girl (7 Choices). These two choices can be
combined, hence total number of choices is:
12
⏟ × ⏟ 7 = 84
𝑩𝒐𝒚 𝑮𝒊𝒓𝒍

𝐽𝑎𝑐𝑘
⏟ , 𝐽𝑖𝑙𝑙
⏟ = 𝐽𝑖𝑙𝑙
⏟ , 𝐽𝑎𝑐𝑘
⏟
𝑀𝑜𝑛𝑖𝑡𝑜𝑟 𝑀𝑜𝑛𝑖𝑡𝑜𝑟 𝑀𝑜𝑛𝑖𝑡𝑜𝑟 𝑀𝑜𝑛𝑖𝑡𝑜𝑟

Part C
There are no restrictions on the gender here. Hence, we have:
19
⏟ × 18
⏟ = 342
𝑴𝒐𝒏𝒊𝒕𝒐𝒓 𝑪𝒐−𝑴𝒐𝒏𝒊𝒕𝒐𝒓

Part D: Direct Counting

From Part B, we know that the number of ways to get two monitors is 84. Out of these, we need to make 1 a
monitor, and the other a co-monitor.
Consider that Jack (boy) and Jill(girl) are two students in the class.
𝐽𝑎𝑐𝑘
⏟ , 𝐽𝑖𝑙𝑙
⏟ ≠ 𝐽𝑖𝑙𝑙
⏟ , 𝐽𝑎𝑐𝑘
⏟
𝑀𝑜𝑛𝑖𝑡𝑜𝑟 𝐶𝑜−𝑚𝑜𝑛𝑖𝑡𝑜𝑟 𝑀𝑜𝑛𝑖𝑡𝑜𝑟 𝐶𝑜−𝑚𝑜𝑛𝑖𝑡𝑜𝑟

Hence, for every pair that was counted in Part B, we need to multiply by 2:
84 × 2 = 168
Part D: Complementary Counting
In part C, we calculated that the number of ways to select a monitor and a co-monitor (with no restrictions)
= 19⏟ × 18
⏟ = 342
𝑴𝒐𝒏𝒊𝒕𝒐𝒓 𝑪𝒐−𝑴𝒐𝒏𝒊𝒕𝒐𝒓

Out of these, subtract the choices where both are boys, or both are girls:
342 − ⏟
12 × 11 − ⏟ 7 × 6 = 342 − 132 − 42 = 168
𝑩𝒐𝒕𝒉 𝑩𝒐𝒚𝒔 𝑩𝒐𝒕𝒉 𝑮𝒊𝒓𝒍𝒔

C. Ordered Pairs (Optional)

In some applications, the order in which the elements are written is important. For example,
➢ the ranking in an exam, or a competition, or a race
➢ distribution of 1st, 2nd and 3rd prizes.
Such pairs are called ordered pairs.

(⏟
𝑎 , ⏟
𝑏 )≠(⏟
𝑏 , ⏟
𝑎 )
𝟏𝒔𝒕 𝟐𝒏𝒅 𝟏𝒔𝒕 𝟐𝒏𝒅

However, if order is not important in a situation. For example, a medical study may be interested in considering
whether a patient survived a medical procedure. In such a case:

( ⏟
𝑎 , ⏟
𝑏 )=( ⏟
𝑏 , ⏟
𝑎 )
𝑺𝒖𝒓𝒗𝒊𝒗𝒆𝒅 𝑺𝒖𝒓𝒗𝒊𝒗𝒆𝒅 𝑺𝒖𝒓𝒗𝒊𝒗𝒆𝒅 𝑺𝒖𝒓𝒗𝒊𝒗𝒆𝒅

By default, the multiplication principle gives us ordered pairs.

Example 4.47: Selecting Monitors

A class has 12 boys and 7 girls. Find the number of ways to choose two monitors.

This looks very similar to Part C of the previous example. There are no restrictions on the gender of the monitor
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➢ Pick the first monitor from any of the 19 students in the class.
➢ For the second monitor, we have only 18 choices, since the first monitor cannot also be the second
monitor.
19
⏟ × 18 ⏟ = 342
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅
𝑴𝒐𝒏𝒊𝒕𝒐𝒓 𝑴𝒐𝒏𝒊𝒕𝒐𝒓
This overcounts the number of ways by two because the way we have counted incorporates:
𝐴𝐵 𝑎𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑓𝑟𝑜𝑚 𝐵𝐴
Hence, the actual number of choices will be:
342
= 171
2

Example 4.48
I need to select a team of two people from a choice of 11 students. How many ways can I do this?

11 × 10
2

4.49: Repetitions Not Allowed

If repetitions are not allowed, the number of objects available for selection will keep decreasing, and this leads
to an especially important pattern:
𝑛 × (𝑛 − 1) × (𝑛 − 2) × …

We have so far seen questions that required us to multiply two numbers.

Now, we see an example where we have more objects.

Example 4.50
I need to select a captain, a vice-captain, and a goalkeeper from a club of eleven people. No person can have
more than one role at a time. Find the number of ways in which the selection can be made.

11
⏟ × 10
⏟ × ⏟
9 = 990
𝐶𝑎𝑝𝑡𝑎𝑖𝑛 𝑽𝒊𝒄𝒆−𝑪𝒂𝒑𝒕𝒂𝒊𝒏 𝑮𝒐𝒂𝒍𝒌𝒆𝒆𝒑𝒆𝒓

D. Repetitions Allowed
The real-life scenarios given to us let us choose certain things:
➢ When selecting two monitors for a class, the two monitors need to be different.
➢ In arranging EXAM, we are arranging letters, and they cannot be repeated.
However, certain scenarios allow for repetition, and this is an important concept.

Example 4.51
How many outfits consisting of a shirt, a pair of trousers, and a pair of shoes can I make if I have five shirts, five
pairs of trousers and five pairs of shoes?

5 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 × ⏟
⏟ 5 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 = 5 × 5 × 5 = 53 = 125
5 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 × ⏟
𝑺𝒉𝒊𝒓𝒕 𝑻𝒓𝒐𝒖𝒔𝒆𝒓𝒔 𝑺𝒉𝒐𝒆𝒔

4.52: Arranging 𝒓 out of 𝒏 objects (Repetition Allowed)

The number of ways of arranging 𝒓 out of 𝒏 objects is:
𝒏
⏟ × 𝒏 ⏟ × …× 𝒏
⏟ = 𝒏𝒓
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝒓𝒕𝒉
𝑳𝒐𝒄𝒂𝒕𝒊𝒐𝒏 𝑳𝒐𝒄𝒂𝒕𝒊𝒐𝒏 𝑳𝒐𝒄𝒂𝒕𝒊𝒐𝒏

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Example 4.53: Situations leading to Repetition

What is the number of possible outcomes if I:
A. Make three-digit numbers using odd digits (if repetition is allowed)
B. Make three-digit numbers using even digits (if repetition is allowed)
C. Answer a question paper with six questions that has four options for each question, if each question
must be answered using one of the given options.
D. Answer a question paper with six questions that has four options for each question, if you have the
option of not attempting one or more of the questions.

Part A
We can structure this counting by thinking of the choices that we have for each digit of the number.
The odd digits are:
1,3, 5, 7, 9 ⇒ 5 𝑂𝑑𝑑 𝐷𝑖𝑔𝑖𝑡𝑠

⏟
5 × ⏟
5 × ⏟
5 = 53 = 125
𝑯𝒖𝒏𝒅𝒓𝒆𝒅′ 𝒔 𝑻𝒆𝒏′ 𝒔 𝑼𝒏𝒊𝒕′ 𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕
Part B
The even digits are:
0,2,4,6,8 ⇒ 5 𝐸𝑣𝑒𝑛 𝐷𝑖𝑔𝑖𝑡𝑠
We have no restrictions on the units digit and the ten’s digit. But zero cannot be the first digit of a three-digit
number.
⏟
4 × ⏟ 5 × ⏟ 5 = 4 × 52 = 100
𝑯𝒖𝒏𝒅𝒓𝒆𝒅′ 𝒔 𝑻𝒆𝒏′ 𝒔 𝑼𝒏𝒊𝒕′ 𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕
Part C
⏟
4 × ⏟
4 × ⏟
4 × ⏟
4 × ⏟
4 × ⏟
4 = 46 = 212 = 4096
𝟏𝒔𝒕 𝟐𝒏𝒅 𝟑𝒓𝒅 𝟒𝒕𝒉 𝟓𝒕𝒉 𝟔𝒕𝒉
𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏
Part D
⏟
5 × ⏟
5 × ⏟
5 × ⏟
5 × ⏟
5 × ⏟
5 = 56
𝟏𝒔𝒕 𝟐𝒏𝒅 𝟑𝒓𝒅 𝟒𝒕𝒉 𝟓𝒕𝒉 𝟔𝒕𝒉
𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏
6
5
E. Complementary Counting

4.54: Complementary Counting

When counting what we do not want is easier than counting what we do want, complementary counting is a
useful technique.

➢ We have already used complementary counting in some of the questions above.

Example 4.55
You are arranging plates for dinner at a dinner table that has numbered seats for four people. Every person gets
exactly one plate to eat from, so you will use 4 plates in all. Plates made from different materials are
distinguishable, but two plates of the same material look identical. What is the number of ways of arranging
plates if you have:
A. 4 ceramic and 4 melamine plates
B. 3 ceramic and 3 melamine plates

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Part A
For the first person, you have a choice of giving him ceramic or melamine, giving you two choices. In fact, for
every seat, you have the same number of choices.
⏟2 × ⏟ 2 × ⏟ 2 × ⏟ 2 = 24 = 16
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑻𝒉𝒊𝒓𝒅 𝑭𝒐𝒖𝒓𝒕𝒉
𝑺𝒆𝒂𝒕 𝑺𝒆𝒂𝒕 𝑺𝒆𝒂𝒕 𝑺𝒆𝒂𝒕
Part B
The only difference between Part A and Part B is that you will not be able to have:
➢ all four seats with ceramic plates OR
➢ all four seats with melamine plates

Hence, the number of choices is:

16 − ⏟
1 − ⏟
1 = 16 − 2 = 14
𝑨𝒍𝒍 𝑭𝒐𝒖𝒓 𝑨𝒍𝒍 𝑭𝒐𝒖𝒓
𝑪𝒆𝒓𝒂𝒎𝒊𝒄 𝑷𝒍𝒂𝒕𝒆𝒔 𝑴𝒆𝒍𝒂𝒎𝒊𝒏𝒆 𝑷𝒍𝒂𝒕𝒆𝒔

F. Further Examples

Example 4.56
I am going to order out from Monday to Friday for lunch. I have a choice from Chinese, Japanese, Indian or
Continental cuisine each day. How many ways can I do this if:
A. I cannot repeat cuisines
B. I can repeat cuisines
C. I can repeat cuisines, but I cannot have the same cuisine all five days of the week. For example,
𝑁𝑜𝑡 𝑉𝑎𝑙𝑖𝑑: 𝐶ℎ𝑖𝑛𝑒𝑠𝑒 + 𝐶ℎ𝑖𝑛𝑒𝑠𝑒 + 𝐶ℎ𝑖𝑛𝑒𝑠𝑒 + 𝐶ℎ𝑖𝑛𝑒𝑠𝑒 + 𝐶ℎ𝑖𝑛𝑒𝑠𝑒
𝑉𝑎𝑙𝑖𝑑: 𝐶ℎ𝑖𝑛𝑒𝑠𝑒 + 𝐶ℎ𝑖𝑛𝑒𝑠𝑒 + 𝐶ℎ𝑖𝑛𝑒𝑠𝑒 + 𝐶ℎ𝑖𝑛𝑒𝑠𝑒 + 𝐽𝑎𝑝𝑎𝑛𝑒𝑠𝑒

Part A: No Restrictions
⏟
4 × ⏟
3 × ⏟
2 × ⏟
1 × ⏟
0 =0
1𝑠𝑡 2𝑛𝑑 3𝑟𝑑 4𝑡ℎ 5𝑡ℎ
𝐷𝑎𝑦 𝐷𝑎𝑦 𝐷𝑎𝑦 𝐷𝑎𝑦 𝐷𝑎𝑦
On the fifth day, no matter what I choose on the earlier four days, I run out of choices.
Hence, I can do this in zero ways.

Part B: No Restrictions
If there are no restrictions
⏟
4 × ⏟
4 × ⏟
4 × ⏟
4 × ⏟
4 = 45 = 210 = 1024
𝑴𝒐𝒏𝒅𝒂𝒚 𝑻𝒖𝒆𝒔𝒅𝒂𝒚 𝑾𝒆𝒅𝒏𝒆𝒔𝒅𝒂𝒚 𝑻𝒉𝒖𝒓𝒔𝒅𝒂𝒚 𝑭𝒓𝒊𝒅𝒂𝒚

Part C
We use complementary counting. The number of ways to order (with no restrictions) is
45 = 1024
The number of ways where the same cuisine gets repeated all five days of the week is:
⏟
1 + ⏟
1 + ⏟1 + ⏟
1 = 4 𝑊𝑎𝑦𝑠
𝐴𝑙𝑙 5 𝐷𝑎𝑦𝑠 𝐴𝑙𝑙 5 𝐷𝑎𝑦𝑠 𝐴𝑙𝑙 5 𝐷𝑎𝑦𝑠 𝐴𝑙𝑙 5 𝐷𝑎𝑦𝑠
𝐶ℎ𝑖𝑛𝑒𝑠𝑒 𝐽𝑎𝑝𝑎𝑛𝑒𝑠𝑒 𝑃𝑢𝑛𝑗𝑎𝑏𝑖 𝐶𝑜𝑛𝑡𝑖𝑛𝑒𝑛𝑡𝑎𝑙

Valid number of ways

1024 − 4 = 1020

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Example 4.57
8 friends deciding between Muggle studies and Arithmancy.
28 = 256

In the above, what if each class needs at least one student?

256 − 2

Example 4.58
Sorting Hat needs to sort 12 twelve students. Number of ways to sort into:
A. Two Houses
B. Three Houses
C. Four Houses

212
312
412

Answer the above if a house must have at least one student

212 − 2
3 − 2(212 )
12

412 − 3[312 − 2(212 )]

Example 4.59
Let 𝑋 = {1,2,3,4,5}. The number of different ordered pairs (𝑌, 𝑍) that can be formed such that 𝑌 ⊆ 𝑋, 𝑍 ⊆ 𝑋, and
𝑌 ∩ 𝑍 is empty is: (JEE Main 2012)

Note that Y and Z are sets.

For each number, we have 3 choices. Total choices
= 35
G. Contrasting arrangements with and without Repetition
Whether repetition is allowed needs to be determined from the question. It can be given either directly, or
indirectly. The number of arrangements with repetition, and without repetition, is, in general, not the same, as
the next example shows.

Example 4.60: Selecting People for Posts

In how many ways can a college committee with twelve professors select a chairperson and a co-chairperson.
Answer the question assuming that a single person can occupy multiple posts. Then, answer again assuming
that a single person can occupy only a single post.

Multiple Posts
Here, one person is allowed to hold multiple posts, and hence, the same person can hold more than one post.
This means that repetition is allowed.
12
⏟ × 12
⏟ = 144
⏟
𝑪𝒉𝒂𝒊𝒓𝒑𝒆𝒓𝒔𝒐𝒏 𝑪𝒐−𝑪𝒉𝒂𝒊𝒓𝒑𝒆𝒓𝒔𝒐𝒏
𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒆 𝑷𝒐𝒔𝒕𝒔 (𝑹𝒆𝒑𝒆𝒕𝒊𝒕𝒊𝒐𝒏 𝑨𝒍𝒍𝒐𝒘𝒆𝒅)
Single Post

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Here, one person is allowed to hold only a single post, and hence, the same person cannot hold more than one
post. This means that repetition is not allowed, and as you choose people for a particular post, the choice of
people will keep reducing.

12
⏟ × 11
⏟ = 132
⏟
𝑪𝒉𝒂𝒊𝒓𝒑𝒆𝒓𝒔𝒐𝒏 𝑪𝒐−𝑪𝒉𝒂𝒊𝒓𝒑𝒆𝒓𝒔𝒐𝒏
𝑫𝒊𝒔𝒕𝒊𝒏𝒄𝒕 𝑷𝒆𝒐𝒑𝒍𝒆 (𝑹𝒆𝒑𝒆𝒕𝒊𝒕𝒊𝒐𝒏 𝑵𝒐𝒕 𝑨𝒍𝒍𝒐𝒘𝒆𝒅)

Example 4.61: Selecting People for Posts

In how many ways can a Reader’s Club with seven members select a President, a Vice-President and a
Treasurer. Answer the question assuming that a single person can occupy multiple posts. Then, answer again
assuming that a single person can occupy only a single post.

Multiple Posts
⏟7 × ⏟
7 × ⏟
7 = 343
⏟
𝑷𝒓𝒆𝒔𝒊𝒅𝒆𝒏𝒕 𝑽𝒊𝒄𝒆−𝑷𝒓𝒆𝒔𝒊𝒅𝒆𝒏𝒕 𝑻𝒓𝒆𝒂𝒔𝒖𝒓𝒆𝒓
𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒆 𝑷𝒐𝒔𝒕𝒔 (𝑹𝒆𝒑𝒆𝒕𝒊𝒕𝒊𝒐𝒏 𝑨𝒍𝒍𝒐𝒘𝒆𝒅)
Single Post
⏟7 × ⏟
6 × ⏟
5 = 210
⏟
𝑷𝒓𝒆𝒔𝒊𝒅𝒆𝒏𝒕 𝑽𝒊𝒄𝒆−𝑷𝒓𝒆𝒔𝒊𝒅𝒆𝒏𝒕 𝑻𝒓𝒆𝒂𝒔𝒖𝒓𝒆𝒓
𝑫𝒊𝒔𝒕𝒊𝒏𝒄𝒕 𝑷𝒆𝒐𝒑𝒍𝒆 (𝑹𝒆𝒑𝒆𝒕𝒊𝒕𝒊𝒐𝒏 𝑵𝒐𝒕 𝑨𝒍𝒍𝒐𝒘𝒆𝒅)

Example 4.62: Arranging Flags

In calculating the number of ways in which red, blue and green flags can be arranged in a line of three, Rohan
did not repeat colours, while Rahul assumed no restrictions. Find the positive difference between Rahul’s
answer and Rohan's Answer.

⏟
3 × ⏟
3 × ⏟
3 − ⏟
3 × ⏟
2 1 = 33 − 6 = 27 − 6 = 21
× ⏟
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑻𝒉𝒊𝒓𝒅 𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑻𝒉𝒊𝒓𝒅
⏟
𝑭𝒍𝒂𝒈 𝑭𝒍𝒂𝒈 𝑭𝒍𝒂𝒈 ⏟
𝑭𝒍𝒂𝒈 𝑭𝒍𝒂𝒈 𝑭𝒍𝒂𝒈
𝑹𝒂𝒉𝒖𝒍′ 𝒔 𝑨𝒏𝒔𝒘𝒆𝒓 𝑹𝒐𝒉𝒂𝒏′ 𝒔 𝑨𝒏𝒔𝒘𝒆𝒓

Example 4.63: Arrangements with Repetition Allowed

Using only the digits 4, 5, 7 and 2, what is the positive difference between the number of three-digit numbers
that can be created if repetition is allowed, and the number of four-digit numbers that can be created if
repetition is not allowed.

𝑇ℎ𝑟𝑒𝑒 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 4 × 4 × 4 = 43 = 64

𝐹𝑜𝑢𝑟 𝐷𝑖𝑔𝑖𝑡 𝑁𝑢𝑚𝑏𝑒𝑟𝑠: 4 × 3 × 2 × 1 = 24
64 − 24 = 40

H. Answering Question Papers

Example 4.64
In how many ways can you attempt the following if you must enter a response:
A. 3 True/False Questions
B. 4 True/False Questions
C. 7 True/False Questions

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23 = 8
24 = 16
27 = 128

Answer the above if not entering a response is also an option?

33 = 27
34 = 81
37

Example 4.65
Multiple Choice Questions (MCQs) need a single option to be marked correct. In how many ways can you
attempt:
A. Questions 1-4 in AMC 12, given that each question has five options
B. Questions 1-5 in the CFA Level I paper, given that each question has 3 options
C. Questions 1-3 in the CAT, given that each question has 4 options

𝐴𝑀𝐶 12: 5 × 5 × 5 × 5 = 54 = 625

𝐶𝐹𝐴: 3 × 3 × 3 × 3 × 3 = 35 = 243
𝐶𝐴𝑇: 4 × 4 × 4 = 43 = 64

Example 4.66
An exam which has three questions, with five answer options each, one of which is to be marked correct for
each question. In how many ways can you
A. Attempt all questions of the paper
B. Attempt two questions of the paper
C. Attempt a single question in the paper
D. Attempt one or more questions of the paper
I. By adding the answers to the earlier three parts
II. By using complementary counting

Part A
5 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 ×
⏟ 5 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
⏟ × ⏟
5 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 = 5 × 5 × 5 = 53 = 125
𝑭𝒊𝒓𝒔𝒕 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑺𝒆𝒄𝒐𝒏𝒅 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑻𝒉𝒊𝒓𝒅 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏

First Second Third

Question Question Question
A A A
B B B
C C C
D D D
E E E

5 5 5 53 = 125

Part B
There are two parts:
➢ Choice of question
✓ Choosing two questions is the same as rejecting one question. Hence, number of ways to choose
questions is 3.
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➢ Choice of option within the question

⏟
3 × 5×5
⏟ = 3 × 25 = 75
𝑪𝒉𝒐𝒊𝒄𝒆 𝒐𝒇 𝑪𝒉𝒐𝒊𝒄𝒆 𝒐𝒇 𝑶𝒑𝒕𝒊𝒐𝒏𝒔
𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏

Part C
There are two parts:
➢ Choice of question
➢ Choice of option within the question
3 × 5 = 15

Part D-I
125
⏟ + 75
⏟ + 15
⏟ = 215
𝑨𝒕𝒕𝒆𝒎𝒑𝒕 𝑨𝒕𝒕𝒆𝒎𝒑𝒕 𝑨𝒕𝒕𝒆𝒎𝒑𝒕
𝟑 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏𝒔 𝟐 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏𝒔 𝟏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏
Part D-II
Total choices in attempting 𝟎 − 𝟑 Questions
Not attempting a question gives an additional choice. Hence, instead of five choices for each question, Pushpak
now has six choices.

Hence, number of Ways to attempt zero or more questions is:

⏟
6 × ⏟
6 × ⏟
6 = 216
𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏
𝑶𝒏𝒆 𝑻𝒘𝒐 𝑻𝒉𝒓𝒆𝒆
Complementary Counting
However, this also includes the “attempt”, where no questions have no choice selected, which we do not want to
count.
Hence, the final answer is
216 − 1 = 215

Example 4.67
A Multiple-Correct, Multiple-Correct question in a JEE paper has question text and four options(𝐴, 𝐵, 𝐶, 𝐷) for
each question. How many different ways are there in which a multiple-correct question can be answered
(keeping in mind that the correct options need to be marked and one or more option(s) is/are correct)? Two
ways of answering the question are {𝐴, 𝐷}{𝐴, 𝐵, 𝐶, 𝐷}.

Multiplication Principle with Repetition

{Option A}
⏟ × {Option B}
⏟ × {Option C}
⏟ × {Option D}
⏟ = 24 = 16
𝟐 𝑪𝒉𝒐𝒊𝒄𝒆𝒔: 𝟐 𝑪𝒉𝒐𝒊𝒄𝒆𝒔: 𝟐 𝑪𝒉𝒐𝒊𝒄𝒆𝒔: 𝟐 𝑪𝒉𝒐𝒊𝒄𝒆𝒔:
𝑪𝒐𝒓𝒓𝒆𝒄𝒕 𝒐𝒓 𝑵𝒐𝒕 𝑪𝒐𝒓𝒓𝒆𝒄𝒕 𝑪𝒐𝒓𝒓𝒆𝒄𝒕 𝒐𝒓 𝑵𝒐𝒕 𝑪𝒐𝒓𝒓𝒆𝒄𝒕 𝑪𝒐𝒓𝒓𝒆𝒄𝒕 𝒐𝒓 𝑵𝒐𝒕 𝑪𝒐𝒓𝒓𝒆𝒄𝒕 𝑪𝒐𝒓𝒓𝒆𝒄𝒕 𝒐𝒓 𝑵𝒐𝒕 𝑪𝒐𝒓𝒓𝒆𝒄𝒕
16 − 1 = 15

Enumeration

{ 𝐴, 𝐵, 𝐶, 𝐷
⏟ ,⏟
𝐴𝐵, 𝐴𝐶, 𝐴𝐷, 𝐵𝐶, 𝐵𝐷, 𝐶𝐷, ⏟
𝐴𝐵𝐶, 𝐴𝐵𝐷, 𝐴𝐶𝐷, 𝐵𝐶𝐷, 𝐴𝐵𝐶𝐷
⏟ }
𝑺𝒊𝒏𝒈𝒍𝒆 𝑪𝒐𝒓𝒓𝒆𝒄𝒕 𝑶𝒑𝒕𝒊𝒐𝒏 𝑻𝒘𝒐 𝑪𝒐𝒓𝒓𝒆𝒄𝒕 𝑶𝒑𝒕𝒊𝒐𝒏𝒔 𝑻𝒉𝒓𝒆𝒆 𝑪𝒐𝒓𝒓𝒆𝒄𝒕 𝑶𝒑𝒕𝒊𝒐𝒏𝒔 𝑭𝒐𝒖𝒓 𝑪𝒐𝒓𝒓𝒆𝒄𝒕 𝑶𝒑𝒕𝒊𝒐𝒏𝒔

Multiplication Principle with Repetition

Think of the multiple choice, multiple correct question as four true\false questions combined, which can be
answered in
24 = 16 𝑤𝑎𝑦𝑠

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16 − 1 = 15

Challenge 4.68 Options Correct Answers

A Matrix-Match question in a JEE paper looks the Match the Statement A Correct Answer I
Column given. It is a multiple-choice, multiple correct question. Statement B Correct Answer II
A statement can have zero or more correct answers, upto the Statement C Correct Answer III
maximum of four mentioned in the question. What is the Statement D Correct Answer IV
number of ways in which the Matrix-Match question above can be answered?

From the previous question, each statement has

24 = 16 𝑐ℎ𝑜𝑖𝑐𝑒𝑠
Each option is to be answered independently, therefore total number of choices is
24 × 24 × 24 × 24 = 216

Example 4.69
Virat is appearing for his midterm exam, which has 9 True/False questions. After he submits the paper, his
teacher informs him that his marks for the True/False questions are neither zero, nor maximum. In how many
ways could Virat have attempted the questions, if (answer each separately):
A. He guessed the answers to all of the questions.
B. He marked the third question as True, and guessed the rest.

29 − 2 = 512 − 2 = 510
28 − 2 = 256 − 2 = 254

Example 4.70 No. of Options No. of Questions True/False

A school asks three types of questions in its Type I Type II Type I Type II Type III
question papers, with Type I and Type II Biology 3 5 4 4 5
multiple choice and Type III as True/False.
Multiple choice questions require a single option to be marked correct. Find the number of ways in which all the
questions of the papers can be attempted.

34 × 54 × 25 = 81 × 10,000 × 2 = 1,620,000

4.4 Counting Numbers and Words-I

A. Idea
Competitions love to ask questions that involve counting numbers. It ticks a number of checkboxes:
➢ Checks understanding of numbers, which students are expected to have, but may not be fluent in
practice.
➢ Allows mixing of concepts from number theory (prime/composite, odd/even, multiples, etc) with
counting concepts. The individual concepts are not very difficult, but the larger the number of concepts
involved, the more difficult the question becomes.
➢ It allows asking of Counting questions in exams in lower grades where it is not officially a part of the
syllabus.

Hence, it is important to practice these questions, till a high degree of comfort is achieved.

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B. Two Digit Numbers

Two-digit numbers are very simple. Yet, they show a lot of underlying structure and properties that can be used
to connect with them with very powerful ideas.
Some of the questions in the example below may be quite easy to solve using one method. Ensure that you are
able to solve them using all the methods given, so that you can use those methods in questions where they are
most suitable.

Example 4.71
Find the number of two-digit numbers with no restrictions.

Multiplication Rule Method

𝑇𝑒𝑛′ 𝑠 𝐷𝑖𝑔𝑖𝑡 ×
⏟ 𝑈𝑛𝑖𝑡 ′ 𝑠 𝐷𝑖𝑔𝑖𝑡
⏟ = 9 × 10 = 90 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟏,𝟐,…,𝟗}: 𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟏,𝟐,…,𝟗}: 𝟏𝟎 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Using Lists
99
⏟ − 10
⏟ + 1 = 90
𝐿𝑎𝑟𝑔𝑒𝑠𝑡 𝑆𝑚𝑎𝑙𝑙𝑒𝑠𝑡
𝑇𝑤𝑜 𝐷𝑖𝑔𝑖𝑡 𝑇𝑤𝑜 𝐷𝑖𝑔𝑖𝑡
𝑁𝑢𝑚𝑏𝑒𝑟 𝑁𝑢𝑚𝑏𝑒𝑟

Example 4.72
Find the number of two-digit numbers which are odd.

Method I: Pair the numbers as follows:

{ (10,11)
⏟ , (12,13)
⏟ , (14,15)
⏟ ,…, (98,99)
⏟ }
𝑭𝒊𝒓𝒔𝒕 𝑷𝒂𝒊𝒓 𝑺𝒆𝒄𝒐𝒏𝒅 𝑷𝒂𝒊𝒓 𝑻𝒉𝒊𝒓𝒅 𝑷𝒂𝒊𝒓 𝑭𝒐𝒓𝒕𝒚 𝑭𝒊𝒇𝒕𝒉 𝑷𝒂𝒊𝒓
Every even number is followed by exactly one odd number.
Hence, the number of odd numbers must be exactly half that of the total two-digit numbers.
90
= = 45
2
Method II: Multiplication Principle
There is no restriction on the ten’s digit.
But the units digit of an odd number must be odd.
𝑇𝑒𝑛′ 𝑠 𝐷𝑖𝑔𝑖𝑡 × ⏟
⏟ 𝑈𝑛𝑖𝑡 ′ 𝑠 𝐷𝑖𝑔𝑖𝑡 = 9 × 5 = 45 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟏,𝟐,…,𝟗}: 𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟏,𝟑,𝟓,𝟕,𝟗}:𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Example 4.73
Find the number of two-digit numbers which are even.

Complementary Counting
Since there are 90 two-digit numbers, and 45 of them are odd, the remaining must be even:
90 − 45 = 45 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟𝑠
Multiplication Principle
𝑇𝑒𝑛′ 𝑠 𝐷𝑖𝑔𝑖𝑡 × ⏟
⏟ 𝑈𝑛𝑖𝑡 ′ 𝑠 𝐷𝑖𝑔𝑖𝑡 = 9 × 5 = 45 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟏,𝟐,…,𝟗}: 𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟐,𝟒,𝟔,𝟖}:𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

4.74: Zero cannot be the first digit: Common Mistake

0 cannot be the first digit of a number, which is two digits or more.
Keep this in mind.

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➢ For example, in the previous question, the number of choices for the ten’s digit is 9 and not 10.

Example 4.75
Find the number of two-digit numbers where both the digits are odd.

The odd digits are:

1,3, 5, 7, 9 ⇒ 5 𝐷𝑖𝑔𝑖𝑡𝑠
′
𝑇𝑒𝑛 𝑠 𝐷𝑖𝑔𝑖𝑡
⏟ × ⏟𝑈𝑛𝑖𝑡 ′ 𝑠 𝐷𝑖𝑔𝑖𝑡 = 5 × 5 = 25 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟏,𝟑,𝟓,𝟕,𝟗}: 𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟏,𝟑,𝟓,𝟕,𝟗}:𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Example 4.76
Find the number of two-digit numbers with exactly two even digits.

The even digits are:

0,2,4,6,8 ⇒ 5 𝐷𝑖𝑔𝑖𝑡𝑠
But we cannot have zero as the first digit, since it will then not be a two-digit number. This reduces the number
of choices for ten’s digit to only 4.
𝑇𝑒𝑛′ 𝑠 𝐷𝑖𝑔𝑖𝑡 × ⏟
⏟ 𝑈𝑛𝑖𝑡 ′ 𝑠 𝐷𝑖𝑔𝑖𝑡 = 4 × 5 = 20 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟐,𝟒,𝟔,𝟖}: 𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟐,𝟒,𝟔,𝟖}:𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Example 4.77
Find the number of two-digit numbers where the digits must be, with repetition allowed, from the set
{3,4,5,6,7,0}.

⏟ ′ 𝑠 𝐷𝑖𝑔𝑖𝑡
𝑇𝑒𝑛 × ⏟ ′ 𝑠 𝐷𝑖𝑔𝑖𝑡
𝑈𝑛𝑖𝑡 = 5 × 6 = 30 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟑,𝟒,𝟓,𝟔,𝟕}: 𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟑,𝟒,𝟓,𝟔,𝟕,𝟎}:𝟔 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Example 4.78
Find the number of two-digit numbers with odd digits, and repetition of digits is not allowed.

𝑇𝑒𝑛′ 𝑠 𝐷𝑖𝑔𝑖𝑡
⏟ 𝑈𝑛𝑖𝑡 ′ 𝑠 𝐷𝑖𝑔𝑖𝑡 = 5 × 4 = 20 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
× ⏟
{𝟏,𝟑,𝟓,𝟕,𝟗}: 𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 (𝟓−𝟏)=𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Example 4.79
Find the number of two-digit numbers with even digits, and repetition of digits, is not allowed.

𝑇𝑒𝑛′ 𝑠 𝐷𝑖𝑔𝑖𝑡 ×
⏟ 𝑈𝑛𝑖𝑡 ′ 𝑠 𝐷𝑖𝑔𝑖𝑡
⏟ = 4 × 4 = 16 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟐,𝟒,𝟔,𝟖}: 𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟐,𝟒,𝟔,𝟖}−𝟏=𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Example 4.80
Find the number of two-digit numbers where the tens digit is greater than three, and the units digit is greater
than five.

⏟ ′ 𝑠 𝐷𝑖𝑔𝑖𝑡
𝑇𝑒𝑛 ⏟ ′ 𝑠 𝐷𝑖𝑔𝑖𝑡 = 6 × 4 = 24 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
× 𝑈𝑛𝑖𝑡
{𝟒,𝟓,𝟔,𝟕,𝟖,𝟗}: 𝟔 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟔,𝟕,𝟖,𝟗}: 𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

4.81: Off by 1 (Common Mistake)

When counting, it is quite easy to count incorrectly, and get your answer different from the actual answer by a
difference of 1.

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➢ For example, in the previous question, you could get

7 × 5 = 35

Example 4.82
Find the number of two-digit numbers with only prime digits.

𝑇𝑒𝑛′ 𝑠 𝐷𝑖𝑔𝑖𝑡 × ⏟
⏟ 𝑈𝑛𝑖𝑡 ′ 𝑠 𝐷𝑖𝑔𝑖𝑡 = 4 × 4 = 16 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟐,𝟑,𝟓,𝟕}: 𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟐,𝟑,𝟓,𝟕}: 𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

C. Choosing the Right Tool

The above set of questions provides a lot of variety regarding different types of two-digit numbers. It also gives
an idea of the power of the multiplication principle.
But this power can also create the impression that this tool should be applied to every question.
In certain cases, it become better or more useful to fall back on a simple enumeration strategy.
Choosing the right method of attack for a question is very important in being able to solve the question. This
comes with practice.
The conditions given can be complicated enough that the final answer is a small number. In such scenarios, it is
often better to use enumeration.

Example 4.83
How many two-digit numbers have ten’s digit exactly two more than the unit’s digit?

Bogus Solution
𝑇𝑒𝑛′ 𝑠 𝐷𝑖𝑔𝑖𝑡 × ⏟
⏟ 𝑈𝑛𝑖𝑡 ′ 𝑠 𝐷𝑖𝑔𝑖𝑡 = 9 × 8 = 72
𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 𝟖 𝑪𝒉𝒐𝒊𝒄𝒆𝒔
This overcounts the numbers because once you fix the Ten’s Digit, with each choice of ten’s digit, there is only
one acceptable choice of Unit’s Digit.

Valid Solution
Pick one number at a time each number as the Ten’s Digit, which makes the number
0𝐴 → 𝑁𝑜𝑡 𝑎 𝑡𝑤𝑜 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟

1𝐴 → 𝑁𝑜 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
2𝐴 → 20
3𝐴 → 31
4𝐴 → 42
5𝐴 → 53
6𝐴 → 64
7𝐴 → 75
8𝐴 → 86
9𝐴 → 97
Shortcut
This can be done using the multiplication principle, creating an elegant (but perhaps not necessarily exam-
suitable) solution:
⏟ ′ 𝑠 𝐷𝑖𝑔𝑖𝑡 ×
𝑇𝑒𝑛 ⏟ ′ 𝑠 𝐷𝑖𝑔𝑖𝑡
𝑈𝑛𝑖𝑡 = 1 × 8 = 8 ⇒ {20,31,42,53,64,75,86,97}
𝑶𝒏𝒍𝒚 𝑶𝒏𝒆 𝑪𝒉𝒐𝒊𝒄𝒆 {𝟎,𝟏,𝟐,𝟑,𝟒,𝟓,𝟔,𝟕}=𝟖 𝑪𝒉𝒐𝒊𝒄𝒆𝒔
𝑨𝒖𝒕𝒐𝒎𝒂𝒕𝒊𝒄𝒂𝒍𝒍𝒚 𝑭𝒊𝒙𝒆𝒅

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Example 4.84: Listing Numbers

How many two-digit numbers have an even tens digit, and the units digit is double of the ten’s digit?

𝑇𝑒𝑛′𝑠 𝐷𝑖𝑔𝑖𝑡 ×
⏟ 𝑈𝑛𝑖𝑡′𝑠 𝐷𝑖𝑔𝑖𝑡
⏟ = 2 × 1 = 2 ⇒ {24,48}
{𝟐,𝟒}: 𝟐 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 𝑵𝒐 𝑪𝒉𝒐𝒊𝒄𝒆=𝟏 𝑪𝒉𝒐𝒊𝒄𝒆

D. Three Digit Numbers

Example 4.85: Three Digit Numbers

Find the number of three-digit numbers which:
A. Have no restrictions
B. Have no zeros (NMTC Primary/Final 2005/02)
C. Are Odd
D. Are multiples of two
E. Have only Odd Digits
F. Have only Even Digits
G. Have only prime digits
H. Have only odd prime digits
I. Have only prime digits and the numbers are odd

Part A
The Hundreds Digit cannot be zero:
𝐻𝑢𝑛𝑑𝑟𝑒𝑑′ 𝑠 𝐷𝑖𝑔𝑖𝑡 × ⏟
⏟ 𝑇𝑒𝑛′ 𝑠 𝐷𝑖𝑔𝑖𝑡 × 𝑈𝑛𝑖𝑡 ′ 𝑠 𝐷𝑖𝑔𝑖𝑡
⏟ = 9 × 10 × 10 = 900 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟏,𝟐,…,𝟗}: 𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟏,𝟐,…,𝟗}: 𝟏𝟎 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟏,𝟐,…,𝟗}: 𝟏𝟎 𝑪𝒉𝒐𝒊𝒄𝒆𝒔
Part B
If we cannot use zero as a digit, then we are left with exactly nine choices for each digit:
𝐻𝑢𝑛𝑑𝑟𝑒𝑑′ 𝑠 𝐷𝑖𝑔𝑖𝑡 × ⏟
⏟ 𝑇𝑒𝑛′ 𝑠 𝐷𝑖𝑔𝑖𝑡 × ⏟ 𝑈𝑛𝑖𝑡 ′ 𝑠 𝐷𝑖𝑔𝑖𝑡 = 9 × 9 × 9 = 729 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟏,𝟐,…,𝟗}: 𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟏,𝟐,…,𝟗}: 𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟏,𝟐,…,𝟗}: 𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔
Part C
900
2
𝐹𝑖𝑟𝑠𝑡 𝐷𝑖𝑔𝑖𝑡
⏟ × 𝑆𝑒𝑐𝑜𝑛𝑑 𝐷𝑖𝑔𝑖𝑡
⏟ × 𝑇ℎ𝑖𝑟𝑑 𝐷𝑖𝑔𝑖𝑡
⏟ = 9 × 10 × 5 = 450 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟏,𝟐,…,𝟗}: 𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟏,𝟐,…,𝟗}: 𝟏𝟎 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟏,𝟑,𝟓,𝟕,𝟗}: 𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔
Part D
𝐹𝑖𝑟𝑠𝑡 𝐷𝑖𝑔𝑖𝑡
⏟ × 𝑆𝑒𝑐𝑜𝑛𝑑 𝐷𝑖𝑔𝑖𝑡
⏟ × 𝑇ℎ𝑖𝑟𝑑 𝐷𝑖𝑔𝑖𝑡
⏟ = 9 × 10 × 5 = 450 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟏,𝟐,…,𝟗}: 𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟏,𝟐,…,𝟗}: 𝟏𝟎 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟐,𝟒,𝟔,𝟖}: 𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Part E
𝐹𝑖𝑟𝑠𝑡 𝐷𝑖𝑔𝑖𝑡
⏟ × ⏟
𝑆𝑒𝑐𝑜𝑛𝑑 𝐷𝑖𝑔𝑖𝑡 × 𝑇ℎ𝑖𝑟𝑑 𝐷𝑖𝑔𝑖𝑡
⏟ = 5 × 5 × 5 = 125 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟏,𝟑,𝟓,𝟕,𝟗}: 𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟏,𝟑,𝟓,𝟕,𝟗}: 𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟏,𝟑,𝟓,𝟕,𝟗}: 𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Part F
𝐹𝑖𝑟𝑠𝑡 𝐷𝑖𝑔𝑖𝑡
⏟ × ⏟
𝑆𝑒𝑐𝑜𝑛𝑑 𝐷𝑖𝑔𝑖𝑡 × 𝑇ℎ𝑖𝑟𝑑 𝐷𝑖𝑔𝑖𝑡
⏟ = 4 × 5 × 5 = 100 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟐,𝟒,𝟔,𝟖}: 𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟐,𝟒,𝟔,𝟖}: 𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟐,𝟒,𝟔,𝟖}: 𝟓 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Part G
Prime digits are:
2,3,5,7 ⇒ 4 𝐷𝑖𝑔𝑖𝑡𝑠
𝐹𝑖𝑟𝑠𝑡 𝐷𝑖𝑔𝑖𝑡
⏟ × ⏟
𝑆𝑒𝑐𝑜𝑛𝑑 𝐷𝑖𝑔𝑖𝑡 × ⏟ 𝑇ℎ𝑖𝑟𝑑 𝐷𝑖𝑔𝑖𝑡 = 4 × 4 × 4 = 64 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟐,𝟑,𝟓,𝟕}: 𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟐,𝟑,𝟓,𝟕}: 𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟐,𝟑,𝟓,𝟕}: 𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

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Part H
Odd Prime digits are:
3,5,7 ⇒ 3 𝐷𝑖𝑔𝑖𝑡𝑠
𝐹𝑖𝑟𝑠𝑡 𝐷𝑖𝑔𝑖𝑡 × ⏟
⏟ 𝑆𝑒𝑐𝑜𝑛𝑑 𝐷𝑖𝑔𝑖𝑡 × ⏟𝑇ℎ𝑖𝑟𝑑 𝐷𝑖𝑔𝑖𝑡 = 3 × 3 × 3 = 27 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟑,𝟓,𝟕}: 𝟑 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟑,𝟓,𝟕}: 𝟑 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟑,𝟓,𝟕}: 𝟑 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

Part I
Prime digits are:
2,3,5,7 ⇒ 4 𝐷𝑖𝑔𝑖𝑡𝑠
𝐹𝑖𝑟𝑠𝑡
⏟ 𝐷𝑖𝑔𝑖𝑡 × 𝑆𝑒𝑐𝑜𝑛𝑑
⏟ 𝐷𝑖𝑔𝑖𝑡 × 𝑇ℎ𝑖𝑟𝑑
⏟ 𝐷𝑖𝑔𝑖𝑡 = 4 × 4 × 3 = 48 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟐,𝟑,𝟓,𝟕}: 𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟐,𝟑,𝟓,𝟕}: 𝟒 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟑,𝟓,𝟕}: 𝟑 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

E. Forming Words

Example 4.86
How many passwords can I frame consisting of three parts such that the first part is a digit from the decimal
system, the second part is a letter from the English alphabet, and the third part is a special character from the
set {!, @, #, $, %, &,∗}.

10
⏟ × 26
⏟ × ⏟
7 = 1820
𝐷𝑖𝑔𝑖𝑡 𝐿𝑒𝑡𝑡𝑒𝑟 𝑆𝑝𝑒𝑐𝑖𝑎𝑙 𝐶ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟

Example 4.872
I want to create a secret language. The language uses the letters {𝐴, 𝐵, 𝐶}, and every word in the language is a
three-letter word. Find the number of possible words if:
A. Repetition is allowed.
B. Repetition is not allowed.
C. Repetition is allowed but the same letter cannot be repeated thrice.

3 × 3 × 3 = 27
3×2×1=6

Complementary Counting
The words where the letters are repeated thrice are:
𝐴𝐴𝐴, 𝐵𝐵𝐵, 𝐶𝐶𝐶 → 3 𝑂𝑝𝑡𝑖𝑜𝑛𝑠
27 − 3 = 24

Example 4.88: Morse Code

Morse code is used in telecommunication and sends signals by means of combination of “dots” and “dashes”. If
you want to encode the letters of the English alphabet, and the digits in the decimal system using a word made
of upto 𝑛 dots/dashes, what is the minimum value of 𝑛?

𝑇𝑜𝑡𝑎𝑙 𝑁𝑜. 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠 = 26 𝐿𝑒𝑡𝑡𝑒𝑟𝑠 + 10 𝐷𝑖𝑔𝑖𝑡𝑠 = 36

The words I can make with length 1 are:

2This is a comparative example. Parts A, B and C denote different conditions that can be applied on the
question. Make you understand the difference.

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{𝐷𝑜𝑡, 𝐷𝑎𝑠ℎ} = 21 = 2

The words I can make with length 2 are:

{𝐷𝑜𝑡 − 𝐷𝑜𝑡, 𝐷𝑎𝑠ℎ − 𝐷𝑎𝑠ℎ, 𝐷𝑜𝑡 − 𝐷𝑎𝑠ℎ, 𝐷𝑎𝑠ℎ − 𝐷𝑜𝑡} = 22 = 4

The words I can make with length 3 are:

23 = 8
The words I can make with length 4 are:
24 = 16
The words I can make with length 5 are:
25 = 32

The total is:

2 + 4 + 8 + 16 + 32 = 62

Example 4.89: Braille Language

A. Braille is a tactile language, which has six dots in a 3 × 2 pattern (three rows and two
columns). Every dot is either raised, or “non-raised”. All letters consists of all six dots. For
example, the diagram alongside represents two raised dots as black, and four regular dots
as white. How many “letters” can you make in the Braille language using the above pattern?
B. The language on which Braille was based used 12 dots. If the same raised/non-raised pattern is used,
how many letters can you make using 12 dots?

Part A
⏟
2 × ⏟
2 2 = 26 = 64
× …× ⏟
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑺𝒊𝒙𝒕𝒉
𝑫𝒐𝒕 𝑫𝒐𝒕 𝑫𝒐𝒕
Part B
212 = 4096

Example 4.90
𝐴𝑛𝑠𝑤𝑒𝑟 𝑒𝑎𝑐ℎ 𝑝𝑎𝑟𝑡 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑒𝑙𝑦
A code maker uses the Greek Letters, Alpha(𝛼), Beta(𝛽), Gamma(𝛾), Delta (Δ), and Chi(𝜒 ) to write a code that
consists of 4 letters. Find the number of possibles codes if:
A. There are no restrictions.
B. Every letter in the code must be unique.
C. The same letter cannot be repeated four times.

Part A
⏟
5 × ⏟
5 × ⏟
5 × ⏟
5 = 54 = 625
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑻𝒉𝒊𝒓𝒅 𝑭𝒐𝒖𝒓𝒕𝒉
𝑳𝒆𝒕𝒕𝒆𝒓 𝑳𝒆𝒕𝒕𝒆𝒓 𝑳𝒆𝒕𝒕𝒆𝒓 𝑳𝒆𝒕𝒕𝒆𝒓
Part B
⏟
5 × ⏟
4 × ⏟
3 × ⏟
2 = 20 × 6 = 120
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑻𝒉𝒊𝒓𝒅 𝑭𝒐𝒖𝒓𝒕𝒉
𝑳𝒆𝒕𝒕𝒆𝒓 𝑳𝒆𝒕𝒕𝒆𝒓 𝑳𝒆𝒕𝒕𝒆𝒓 𝑳𝒆𝒕𝒕𝒆𝒓
Part C
625
⏟ − ⏟
5 = 620
𝑻𝒐𝒕𝒂𝒍 𝑹𝒆𝒑𝒆𝒂𝒕𝒆𝒅
𝑪𝒉𝒐𝒊𝒄𝒆𝒔 𝑭𝒐𝒖𝒓 𝑻𝒊𝒎𝒆𝒔

Example 4.91
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𝐴𝑛𝑠𝑤𝑒𝑟 𝑒𝑎𝑐ℎ 𝑝𝑎𝑟𝑡 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑒𝑙𝑦

A language has exactly four words 𝑘𝑎𝑦, 𝑘𝑦𝑎, 𝑎𝑘𝑦, and 𝑦𝑘𝑎. Each sentence in the language consists of four words.
Find the number of possible sentences if:
A. the same word cannot be immediately repeated. For example, 𝑘𝑎𝑦 𝑘𝑦𝑎 𝑘𝑎𝑦 𝑘𝑦𝑎 is a valid sentence but
𝑘𝑎𝑦 𝑘𝑎𝑦 𝑘𝑦𝑎 𝑦𝑘𝑎 is not.
B. Sentences must be written using words in either ascending order (alphabetically), or descending order
(alphabetically).
C. If every word can be written in either capital letters (𝐾𝐴𝑌) or small letters (𝑘𝑎𝑦), and the two are
considered different.
D. If every letter can be written in either capital letters (𝐾) or small letters (𝑘), and 𝑘𝐴𝑦 is different from
𝑘𝑎𝑦.

Part A
⏟
4 × ⏟
3 × ⏟
3 × ⏟
3 = 108 𝑆𝑒𝑛𝑡𝑒𝑛𝑐𝑒𝑠
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑻𝒉𝒊𝒓𝒅 𝑭𝒐𝒖𝒓𝒕𝒉
𝑾𝒐𝒓𝒅 𝑾𝒐𝒓𝒅 𝑾𝒐𝒓𝒅 𝑾𝒐𝒓𝒅
Part B
If you arrange the words in ascending order, you get one sentence:
𝑎𝑘𝑦, 𝑘𝑎𝑦, 𝑘𝑦𝑎, 𝑦𝑘𝑎
If you arrange the words in descending order, you get one sentence:
𝑦𝑘𝑎, 𝑘𝑦𝑎, 𝑘𝑎𝑦, 𝑎𝑘𝑦
The total number is:
1+1=2
Part C
You get 8 choices for each word:
⏟
8 × ⏟ 8 × ⏟ 8 × ⏟ 8 = 84 = 212 = 4096
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑻𝒉𝒊𝒓𝒅 𝑭𝒐𝒖𝒓𝒕𝒉
𝑾𝒐𝒓𝒅 𝑾𝒐𝒓𝒅 𝑾𝒐𝒓𝒅 𝑾𝒐𝒓𝒅
Part D
Each word can be written in
⏟
2 × ⏟
2 × ⏟
2 = 8 𝑊𝑎𝑦𝑠
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑻𝒉𝒊𝒓𝒅
𝑳𝒆𝒕𝒕𝒆𝒓 𝑳𝒆𝒕𝒕𝒆𝒓 𝑳𝒆𝒕𝒕𝒆𝒓
Since every word given in the question can be written in 8 ways, the number of “words” that we have is:
4 × 8 = 32 𝑊𝑜𝑟𝑑𝑠
Hence, the number of sentences is:
32 × 32 × 32 × 32 = 324 = (25 )4 = 220

Example 4.92
𝐴𝑛𝑠𝑤𝑒𝑟 𝑒𝑎𝑐ℎ 𝑝𝑎𝑟𝑡 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑒𝑙𝑦
How many 5-words sentences can I make in a language that consists of exactly three words: “brillig”, “slithy”,
and “toves” if:
A. There are no restrictions.
B. A sentence cannot consist of the same word repeated five times.

Part A
⏟
3 × ⏟
3 × ⏟
3 × ⏟
3 3 = 35 = 243
× ⏟
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑻𝒉𝒊𝒓𝒅 𝑭𝒐𝒖𝒓𝒕𝒉 𝑭𝒊𝒗𝒆
𝑾𝒐𝒓𝒅 𝑾𝒐𝒓𝒅 𝑾𝒐𝒓𝒅 𝑾𝒐𝒓𝒅 𝑾𝒐𝒓𝒅
Part B
The number of sentences with the same word repeated five times are:
𝑏𝑟𝑖𝑙𝑙𝑖𝑔, 𝑏𝑟𝑖𝑙𝑙𝑖𝑔, 𝑏𝑟𝑖𝑙𝑙𝑖𝑔, 𝑏𝑟𝑖𝑙𝑙𝑖𝑔, 𝑏𝑟𝑖𝑙𝑙𝑖𝑔

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𝑠𝑙𝑖𝑡ℎ𝑦, 𝑠𝑙𝑖𝑡ℎ𝑦, 𝑠𝑙𝑖𝑡ℎ𝑦, 𝑠𝑙𝑖𝑡ℎ𝑦, 𝑠𝑙𝑖𝑡ℎ𝑦

𝑡𝑜𝑣𝑒𝑠, 𝑡𝑜𝑣𝑒𝑠, 𝑡𝑜𝑣𝑒𝑠, 𝑡𝑜𝑣𝑒𝑠, 𝑡𝑜𝑣𝑒𝑠

Using complementary counting, the number of possible sentences is:

243
⏟ − ⏟
3 = 240
𝑻𝒐𝒕𝒂𝒍 𝑹𝒆𝒑𝒆𝒂𝒕𝒆𝒅
𝑺𝒆𝒏𝒕𝒆𝒏𝒄𝒆𝒔 𝑭𝒊𝒗𝒆 𝑻𝒊𝒎𝒆𝒔

Example 4.93
Find the number of words (not necessarily meaningful) made from letters in the English Alphabet if:
A. Each word has two letters, and repetition is allowed
B. Each word has two letters, and repetition is not allowed
C. Each word has two letters, of which one is a vowel, and the other is a consonant.

Part A
I have twenty-six choices for the first letter. Similarly, I have twenty-six choices for the second letter.
The choices are independent. That is, I can mix the choices in any way I want.

Hence, the number of words I can make is:

26
⏟ × 26
⏟ = 676
𝑭𝒊𝒓𝒔𝒕 𝑳𝒆𝒕𝒕𝒆𝒓 𝑺𝒆𝒄𝒐𝒏𝒅 𝑳𝒆𝒕𝒕𝒆𝒓
Part B
I have twenty-six choices for the first letter. Similarly, I have twenty-six choices for the second letter, but I
cannot the repeat the letter which was used first, and hence the number of choices comes down to 25.

The choices are independent. That is, I can mix the choices in any way I want
26
⏟ × 25
⏟ = 650
𝑭𝒊𝒓𝒔𝒕 𝑳𝒆𝒕𝒕𝒆𝒓 𝑺𝒆𝒄𝒐𝒏𝒅 𝑳𝒆𝒕𝒕𝒆𝒓

Part C
First, we need to decide the order of the vowel and the consonant. There are two choices:
➢ Vowel followed by consonant
➢ Consonant followed by vowel

And we need to find the number of choices in each:

5⏟× 21 + 21
⏟×5 = 105 + 105 = 210
𝑽𝒐𝒘𝒆𝒍+𝑪𝒐𝒏𝒔𝒐𝒏𝒂𝒏𝒕 𝑪𝒐𝒏𝒔𝒐𝒏𝒂𝒏𝒕+𝑽𝒐𝒘𝒆𝒍

Example 4.94
How many two-letters words in the English language made of consonants have at least one 𝐵. (For this
question, consider 𝑌 to be a vowel).

Complementary Counting
𝑇𝑜𝑡𝑎𝑙 𝑊𝑜𝑟𝑑𝑠 − 𝑊𝑜𝑟𝑑𝑠 𝑤𝑖𝑡ℎ 𝑁𝑜 𝐵 = 202 − 192 = 400 − 361 = 39

Direct Counting

First Letter is a B
1 × 20 = 20
Second Letter is a B

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20 × 1 = 20
Both Letters are B
1×1=1

20 + 20 − 1 = 40 − 1 = 39

Example 4.95
In a certain land, they really like the number four. The alphabet has only four letters, and their words have a
maximum length of four. Since they have so few choices, they make their letters work really hard. They use all
the words that can be made. How many words can be made?

We can divide this into four cases:

➢ Words of Length 1
➢ Words of Length 2
➢ Words of Length 3
➢ Words of Length 3

Add up the number of words in each to get the total number of words.

⏟
4 + ⏟2
4 + ⏟3
4 + ⏟4
4 = 4 + 16 + 64 + 256 = 340
𝑳𝒆𝒏𝒈𝒕𝒉:𝟏 𝑳𝒆𝒏𝒈𝒕𝒉:𝟐 𝑳𝒆𝒏𝒈𝒕𝒉:𝟑 𝑳𝒆𝒏𝒈𝒕𝒉:𝟒

Example 4.96
Phi(𝜙), Tau(𝜏), and Chi(𝜒) are three letters of the Greek alphabet. How many (not necessarily meaningful)
words of maximum length four can you make which use the letter 𝜙 at least once?

Complementary Counting
Here, we need to complementary counting. We will count the total number of words, and from that we will
subtract the words which do not use the letter 𝜙 even once.
However, the question asks for words of maximum length four, and hence, we need to break the problem down
into cases.
Casework with Complementary Counting
𝑊𝑜𝑟𝑑𝑠 𝑤𝑖𝑡ℎ 𝜙 = 𝑇𝑜𝑡𝑎𝑙 𝑊𝑜𝑟𝑑𝑠 − 𝑊𝑜𝑟𝑑𝑠 𝑤𝑖𝑡ℎ 𝑛𝑜 𝜙

Length of Total Words Words with No 𝜙 Words with at

Word least one 𝜙
1 3 2 1
2 32 = 9 22 = 4 5
3 33 = 27 23 = 8 19
4 34 = 81 24 = 16 65

120 30 90

4.5 Difficult Questions on Counting

A. Multiple Restrictions
Certain questions will impose multiple conditions on what is to be counted. Such questions have to be handled
carefully. They may be more difficult when done one way, and much easier when done another way. The best
way to learn how to do such questions is by practice.

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4.97: Multiple Restrictions

When encountering multiple restrictions, it often makes sense to handle the most restrictive condition first.

➢ This is a suggestion, not a rule.

Example 4.98
How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and
all four digits are different? (AMC 8 1995/23)

𝐿𝑒𝑓𝑡𝑚𝑜𝑠𝑡 𝑑𝑖𝑔𝑖𝑡 = 𝑂𝑑𝑑 = 5 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

𝑆𝑒𝑐𝑜𝑛𝑑 𝐷𝑖𝑔𝑖𝑡 = 𝐸𝑣𝑒𝑛 = 5 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
𝑇ℎ𝑖𝑟𝑑 𝐷𝑖𝑔𝑖𝑡 = 10 − 2 = 8
4𝑡ℎ 𝐷𝑖𝑔𝑖𝑡 = 10 − 3 = 8
⏟
5 × ⏟
5 × ⏟ 8 × ⏟ 7 = 200 × 7 = 1400
𝑻𝒉𝒐𝒖𝒔𝒂𝒏𝒅𝒔 𝑯𝒖𝒏𝒅𝒓𝒆𝒅𝒔 𝑻𝒆𝒏𝒔 𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕

B. Complementary Counting

Example 4.99
A. How many whole numbers between 1 and 1000 do not contain the digit 1? (AMC 8 2009/22)
B. How many whole numbers less than 1000 have at least the digit 1 at least once?

Part A
We use all digits other the digit 1.
{0,2,3,4,5,6,7,8,9} ⇒ 9 𝐷𝑖𝑔𝑖𝑡𝑠
We are left 9 choices for each place. We would normally exclude zero from the counting, but here we are
counting 1-digit, 2-digit and 3-digit numbers, which means we can include zero:
𝐻𝑢𝑛𝑑𝑟𝑒𝑑′𝑠
⏟ 𝐷𝑖𝑔𝑖𝑡 × ⏟ 𝑇𝑒𝑛′𝑠 𝐷𝑖𝑔𝑖𝑡 × ⏟ 𝑈𝑛𝑖𝑡′𝑠 𝐷𝑖𝑔𝑖𝑡 = 9 × 9 × 9 = 729 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
{𝟎,𝟐,…,𝟗}: 𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟐,…,𝟗}: 𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 {𝟎,𝟐,…,𝟗}: 𝟗 𝑪𝒉𝒐𝒊𝒄𝒆𝒔

But the 729 number also include 0, which we don’t want to count. Hence, the final answer is:
729 − 1 = 728
Part B
There are 1000 numbers in the range
0,1,2, … ,999
Of these, 728 do not contain the digit 1. Hence, the number of numbers which do contain the digit 1 are
999 − 728 = 271

Example 4.100
𝟑𝟔 𝟑 𝟏 and 𝟔𝟏 are four physical blocks. Using these four blocks, how many six-digit numbers can be
formed? (NMTC Final/Primary 2005/4)

We can arrange these four blocks in:

⏟
4 × ⏟
3 × ⏟
2 × ⏟
1 = 24 𝑤𝑎𝑦𝑠
𝑭𝒊𝒓𝒔𝒕 𝑩𝒍𝒐𝒄𝒌 𝑺𝒆𝒄𝒐𝒏𝒅 𝑩𝒍𝒐𝒄𝒌 𝑻𝒉𝒊𝒓𝒅 𝑩𝒍𝒐𝒄𝒌 𝑭𝒐𝒖𝒓𝒕𝒉 𝑩𝒍𝒐𝒄𝒌
Out of the 24 arrangements, two arrangements will give the same number:
𝟑𝟔 𝟏 𝟑 𝟔𝟏 and 𝟑 𝟔𝟏 𝟑𝟔 𝟏
Hence, the total number of six-digit numbers is:

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24 − 1 = 23
C. Casework
In general, questions which need to be handled using cases require far more effort and care then regular
questions. Some things to take care of:
➢ Cases need to be identified carefully
➢ Cases should not overlap (𝑚𝑢𝑡𝑢𝑎𝑙𝑙𝑦 𝑒𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒).
➢ Cases should together cover the entire requirement that you want to count (𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦 𝑒𝑥ℎ𝑎𝑢𝑠𝑡𝑖𝑣𝑒).

Example 4.101
How many three-digit even numbers greater than 300, with distinct digits, can be formed from the set
{0,1,3,5,8}.

Step I: First Digit Case I: First Digit is 8

Since the number is greater than 300, the first digit Last digit must be 0.
cannot be zero or one. Hence, the first digit must be 3 choices for middle digit: 1,3,5
one out of 810,830,850
3, 5, 𝑜𝑟 8 Case II: First Digit is Odd
Last digit can be 0 or 8: 2 Choices
Step II: Last Digit Middle Digit: 3 Choices
The possible numbers for the last digit are 0 and 8. 𝑇𝑜𝑡𝑎𝑙 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 = ⏟ 2 × ⏟ 3 × ⏟
2 = 12
At this stage, it is important to recognize that if 8 is 𝐹𝑖𝑟𝑠𝑡 𝑀𝑖𝑑𝑑𝑙𝑒 𝐿𝑎𝑠𝑡
𝐷𝑖𝑔𝑖𝑡 𝐷𝑖𝑔𝑖𝑡 𝐷𝑖𝑔𝑖𝑡
the first digit, then it cannot be the last digit. The total number of choices is
Hence, we need to consider two different cases: = 12 + 3 = 15

Example 4.102
How many whole numbers between 99 and 999 contain exactly one 0? (AMC 8 2002/19)

Method I
Numbers greater than 99 and less than 999 must be three digits numbers. The zero cannot be in the hundred’s
place. Hence, we consider two cases:

𝑍𝑒𝑟𝑜 𝑖𝑛 𝑈𝑛𝑖𝑡𝑠 𝐷𝑖𝑔𝑖𝑡: ⏟

9 × ⏟
9 = 81
𝑯𝒖𝒏𝒅𝒓𝒆𝒅𝒔 𝑻𝒆𝒏𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕
𝑍𝑒𝑟𝑜 𝑖𝑛 𝑇𝑒𝑛𝑠 𝐷𝑖𝑔𝑖𝑡: ⏟
9 × ⏟
9 = 81
𝑯𝒖𝒏𝒅𝒓𝒆𝒅𝒔 𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕
𝑇𝑜𝑡𝑎𝑙 = 81 + 81 = 162
Method II
We can also subtract the numbers with no zeros, and the numbers with two zeros from the total number of
three digit numbers:
900
⏟ − 9 ⏟3 − ⏟ 9 = 900 − 729 − 9 = 162
𝑻𝒉𝒓𝒆𝒆 𝑵𝒐 𝑻𝒘𝒐
𝑫𝒊𝒈𝒊𝒕 𝑵𝒖𝒎𝒃𝒆𝒓𝒔 𝒁𝒆𝒓𝒐𝒆𝒔 𝒁𝒆𝒓𝒐𝒆𝒔

Example 4.103
The number of nonnegative integers which are less than 1000, and end with only one zero is: (NMTC Sub-
Junior/Screening 2004/8)

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𝑇𝑒𝑛′ 𝑠 𝐷𝑖𝑔𝑖𝑡 𝑖𝑠 𝑁𝑜𝑡 𝑍𝑒𝑟𝑜: 10

⏟ × ⏟
9 × ⏟
1 = 10 × 9 × 1 = 90
𝑯𝒖𝒏𝒅𝒓𝒆𝒅 𝑻𝒆𝒏′ 𝒔 𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕
𝑇𝑒𝑛′ 𝑠 𝐷𝑖𝑔𝑖𝑡 𝑖𝑠 𝑍𝑒𝑟𝑜: 𝑂𝑛𝑙𝑦 𝑂𝑛𝑒 𝑁𝑢𝑚𝑏𝑒𝑟: 0
𝑇𝑜𝑡𝑎𝑙 = 90 + 1 = 91
Complementary Counting
We want to the last digit to be a zero. Consider the numbers less than 1000 that end with a zero:
0,10,20,30, … ,990
If we consider only the first two digits we get:
0,1,2, … ,99 ⇒ 100 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
But we do not want the numbers with first two digits:
10,20, … ,90 ⇒ 9 𝑁𝑢𝑚𝑏𝑒𝑟𝑠
And hence the final answer is
100 − 9 = 91

D. Divisibility

Example 4.104
The total number of possible proper three-digit integers that can be formed using 0,1,3,4 and 5 without
repetition such that they are divisible by 5 are: (JMET 2011/69)

Since the number must be divisible by 5, the last digit must be either 5 or 0.

Case I: The last digit is 0:

4 × 3 × 1 = 12

Case II: The last digit is 5:

3×3×1=9
Total
= 12 + 9 = 21

Example 4.105
Three spinners are shown. The spinners are used to determine the
hundreds, tens and ones digits of a three-digit number. How many
possible three-digit numbers that can be formed in this way are
divisible by 6? (Gauss 8 2020/22)

Number is divisible by 6 if and only if it is divisible by 2 and 3. Valid one’s digits are only 0 and 2.

If the one’s digit is 2, then the other two digits should be two less than a multiple of 3:
𝑇𝑜𝑡𝑎𝑙 = 7 = (1,6), (2,5)
𝑇𝑜𝑡𝑎𝑙 = 10 = (2,8), (3,7), (4,6)
5 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

If one’s digit is 0, then the other two numbers have a total which is a multiple of 3:
𝑇𝑜𝑡𝑎𝑙 = 6 = (1,5)
𝑇𝑜𝑡𝑎𝑙 = 9 = (1,8), (2,7), (3,6), (4,5)
𝑇𝑜𝑡𝑎𝑙 = 12 = (4,8)
6 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

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E. Exponents
When the number of choices that we get repeats, then the expression that we get when counting is best
expressed using exponents.

Example 4.106
Assume every 7-digit whole number is a possible telephone number except those that begin with 0 or 1. What
fraction of telephone numbers begin with 9 and end with 0? (AMC 8 1985/22)

Total Telephone Numbers:

First Digit cannot be zero or one. Hence, we are left with 8 digits for the first digit.
⏟
8 × 10
⏟ × 10
⏟ × 10
⏟ × 10
⏟ × 10
⏟ × 10
⏟ = 8 × 106
𝑭𝒊𝒓𝒔𝒕 𝑫𝒊𝒈𝒊𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑫𝒊𝒈𝒊𝒕 𝑻𝒉𝒊𝒓𝒅 𝑫𝒊𝒈𝒊𝒕 𝑭𝒐𝒖𝒓𝒕𝒉 𝑫𝒊𝒈𝒊𝒕 𝑭𝒊𝒇𝒕𝒉 𝑫𝒊𝒈𝒊𝒕 𝑺𝒊𝒙𝒕𝒉 𝑫𝒊𝒈𝒊𝒕 𝑺𝒆𝒗𝒆𝒏𝒕𝒉 𝑫𝒊𝒈𝒊𝒕
Eligible Telephone Numbers
Number of telephone numbers that begin with 9, and end with zero:
⏟
1 × 10
⏟ × 10
⏟ × 10
⏟ × 10
⏟ × 10
⏟ × ⏟
1 = 105
𝑭𝒊𝒓𝒔𝒕 𝑫𝒊𝒈𝒊𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑫𝒊𝒈𝒊𝒕 𝑻𝒉𝒊𝒓𝒅 𝑫𝒊𝒈𝒊𝒕 𝑭𝒐𝒖𝒓𝒕𝒉 𝑫𝒊𝒈𝒊𝒕 𝑭𝒊𝒇𝒕𝒉 𝑫𝒊𝒈𝒊𝒕 𝑺𝒊𝒙𝒕𝒉 𝑫𝒊𝒈𝒊𝒕 𝑺𝒆𝒗𝒆𝒏𝒕𝒉 𝑫𝒊𝒈𝒊𝒕

105 1
Required Fraction = 6
=
8 × 10 80
Shortcut
The middle digits have the same choice in both the fractions. Hence, the comparison is only in the first digit and
the last digit, where we have:
1×1 1
=
8 × 10 80
F. Palindromes
A palindrome is a word that reads the same backwards and forwards. A palindrome can be a number also.
For example:
➢ 22 is a two-digit palindrome
➢ 343 is a three-digit palindrome
➢ 3553 is a four-digit palindrome

Warmup 4.107
Find the:
1. Largest three-digit palindrome that does not have all digits the same
2. Smallest four-digit palindrome that does not have all digits the same

989
1001

Example 4.108
Determine the form that palindromes from one to five digits must have (answer for each separately). Use letters
for digits.

⏟
𝑎 , 𝑎𝑎
⏟ , 𝑎𝑏𝑎
⏟ , ⏟
𝑎𝑏𝑏𝑎 , 𝑎𝑏𝑐𝑏𝑎
⏟
𝑶𝒏𝒆 𝑫𝒊𝒈𝒊𝒕 𝑻𝒘𝒐 𝑫𝒊𝒈𝒊𝒕 𝑻𝒉𝒓𝒆𝒆 𝒅𝒊𝒈𝒊𝒕 𝑭𝒐𝒖𝒓 𝒅𝒊𝒈𝒊𝒕 𝑭𝒊𝒗𝒆 𝒅𝒊𝒈𝒊𝒕
𝑷𝒂𝒍𝒊𝒏𝒅𝒓𝒐𝒎𝒆 𝑷𝒂𝒍𝒊𝒏𝒅𝒓𝒐𝒎𝒆 𝑷𝒂𝒍𝒊𝒏𝒅𝒓𝒐𝒎𝒆 𝑷𝒂𝒍𝒊𝒏𝒅𝒓𝒐𝒎𝒆 𝑷𝒂𝒍𝒊𝒏𝒅𝒓𝒐𝒎𝒆

Where

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𝑎, 𝑏, 𝑐 𝑎𝑟𝑒 𝑑𝑖𝑔𝑖𝑡𝑠

Example 4.109
How many palindromes one or greater are there which are:
A. One Digit
B. Two Digits
C. Three Digits
D. Four Digits
E. Five Digits

One Digit
There is no restriction on numbers of one digit. They are all palindromes.
⏟
9 =9
𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕
Two Digit
Two-digit palindromes are of the form
𝑎𝑎
The units digit has to be the same as the ten’s digit. Hence, we only have a choice in the ten’s digit.
⏟
9 × ⏟ 1 =9
𝑻𝒆𝒏𝒔 𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕
Three Digit
Three-digit palindromes are of the form
𝑎𝑏𝑎
Once we choose the hundred’s digit, the one’s digit is automatically chosen, leaving us with no choices there.
We can choose any digit from 0-9 in the ten’s digit, giving us ten choices.

⏟
9 × 10
⏟ × ⏟
1 = 9 × 10 = 90
𝑯𝒖𝒏𝒅𝒓𝒆𝒅𝒔 𝑻𝒆𝒏𝒔 𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕
Four Digit
Four-digit palindromes are of the form
𝑎𝑏𝑏𝑎
Once we choose the thousand’s digit, the one’s digit is automatically chosen, leaving us with no choices there.
We can choose any digit from 0-9 in the hundred’s digit, which leaves us with no choices in the ten’s digit.

⏟
9 × 10
⏟ × ⏟
1 × ⏟
1 = 90
𝑻𝒉𝒐𝒖𝒔𝒂𝒏𝒅𝒔 𝑯𝒖𝒏𝒅𝒓𝒆𝒅𝒔 𝑻𝒆𝒏𝒔 𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕

Part E: Five Digits

⏟
9 × 10
⏟ × 10
⏟ × ⏟
1 × ⏟
1 = 900
𝑻𝒆𝒏 𝑻𝒉𝒐𝒖𝒔𝒂𝒏𝒅𝒔 𝑻𝒉𝒐𝒖𝒔𝒂𝒏𝒅𝒔 𝑯𝒖𝒏𝒅𝒓𝒆𝒅𝒔 𝑻𝒆𝒏𝒔 𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕

4.1: Number of 𝒏 digit Palindromes

𝑵𝒐.𝒐𝒇 𝑫𝒊𝒈𝒊𝒕𝒔
⌈ 𝟐 ⌉−𝟏
𝟗 × 𝟏𝟎

Where ⌈𝑥⌉ represents the ceiling function, which is the smallest integer less than or equal to 𝑥.

Example 4.110
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Find the number of

A. Five-digit palindromes
B. Six-digit palindromes
C. Nine-digit palindromes
D. 100-digit Palindromes
E. 101-Digit Palindromes

5
⌈ ⌉−1
9 × 10 2 = 9 × 10⌈2.5⌉−1 = 9 × 103−1 = 9 × 102 = 9 × 100 = 900

6
⌈ ⌉−1
9 × 10 2 = 9 × 10⌈3⌉−1 = 9 × 103−1 = 9 × 102 = 9 × 100 = 900

9
⌈ ⌉−1
9 × 10 2 = 9 × 10⌈4.5⌉−1 = 9 × 105−1 = 9 × 104 = 9 × 10000 = 90000

100
⌈ ⌉−1
9 × 10 2 = 9 × 10⌈50⌉−1 = 9 × 1049

101
⌈ ⌉−1
9 × 10 2 = 9 × 10⌈50.5⌉−1 = 9 × 1051−1 = 9 × 1050

Example 4.111
How many three-digit multiples of five are palindromes?

If the number is a multiple of 5, the last digit must be

0 𝑜𝑟 5
But, the first digit must be the same as the last digit. Hence, the last digit cannot be 0.

Hence, the number must be of the form

5𝑋5 → 𝑤ℎ𝑒𝑟𝑒 𝑋 𝑖𝑠 𝑠𝑜𝑚𝑒 𝑑𝑖𝑔𝑖𝑡
We can put 10 digits in place of X, giving us 10 numbers:
505,515, … ,595 ⇒ 10 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

Answer the above question for four digits.

As above, the last digit has to be 5.

Also, the moment we choose the hundred’s digit, the ten’s digit is automatically chosen.

Hence, the numbers must be:

5005,5115, … 5995 ⇒ 10 𝑁𝑢𝑚𝑏𝑒𝑟𝑠

Example 4.112
What is the number of even three-digit palindromes?

If the number is a multiple of 2, the last digit must be

0, 2,4, 6, 8
But, the first digit must be the same as the last digit. Hence, the last digit cannot be 0.

Hence, we must have:

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⏟
1 × 10
⏟ × ⏟
4 = 10 × 4 = 40
𝑯𝒖𝒏𝒅𝒓𝒆𝒅𝒔 𝑻𝒆𝒏𝒔 𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕

Answer the above question for four digits.

If the number is a multiple of 2, the last digit must be

0, 2,4, 6, 8
But, the first digit must be the same as the last digit. Hence, the last digit cannot be 0.

Also, the moment we choose the hundred’s digit, the ten’s digit is automatically chosen.

Hence, we must have:

⏟
1 × 10
⏟ × ⏟
1 × ⏟
4 = 10 × 4 = 40
𝑻𝒉𝒐𝒖𝒔𝒂𝒏𝒅𝒔 𝑯𝒖𝒏𝒅𝒓𝒆𝒅𝒔 𝑻𝒆𝒏𝒔 𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕

Example 4.113
A palindrome is a number that reads the same forward and backwards. How many positive 3-digit palindromes
are multiples of 3? (Mathcounts 2003 Warm-up 7)

Strategy
A three-digit palindrome must be of the form:
𝑎𝑏𝑎
We do this using casework on the first digit, since deciding the first digit also decides the last digit.

Casework

Sum of first digit Possible Middle Numbers No. of

and last digit Digits Numbers
1_1 2 1,4,7 111 3
141
171
2_2 4 2,5,8 222 3
252
282
3_3 6 0,3,6,9 303 4 10
333
363
393
4_4 8 1,4,7 414 3
444
474
5_5 10 2,5,8 525 3
555
585
6_6 12 0,3,6,9 606 4 10
636
666
696
7_7 14 1,4,7 717 3
747

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777
8_8 16 2,5,8 828 3
858
888
9_9 18 0,3,6,9 909 4 10
939
969
999
30

Example 4.114
How many palindromes are there, which:
A. Have three digits, all of which are odd
B. Have four digits, all of which are prime
C. Have five digits, all digits are prime, and the number itself is odd

Part A
Choices for the hundred’s digit:
1,3,5,7,9 ⇒ 5 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
⏟
5 × ⏟ 5 × ⏟ 1 = 25
𝑯𝒖𝒏𝒅𝒓𝒆𝒅𝒔 𝑻𝒆𝒏𝒔 𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕
Part B
Choices for prime digit:
2,3,5,7 ⇒ 4 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
⏟
4 × ⏟
4 × ⏟1 × ⏟ 1 = 4 × 4 = 16
𝑻𝒉𝒐𝒖𝒔𝒂𝒏𝒅𝒔 𝑯𝒖𝒏𝒅𝒓𝒆𝒅𝒔 𝑻𝒆𝒏𝒔 𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕
Part C
Choices for odd prime digit:
3,5,7 ⇒ 3 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
Choices for prime digit:
2,3,5,7 ⇒ 4 𝐶ℎ𝑜𝑖𝑐𝑒𝑠

⏟
3 × ⏟
4 × ⏟
4 × ⏟
1 × ⏟
1 = 48
𝑻𝒆𝒏 𝑻𝒉𝒐𝒖𝒔𝒂𝒏𝒅𝒔 𝑻𝒉𝒐𝒖𝒔𝒂𝒏𝒅𝒔 𝑯𝒖𝒏𝒅𝒓𝒆𝒅𝒔 𝑻𝒆𝒏𝒔 𝑼𝒏𝒊𝒕𝒔
𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕 𝑫𝒊𝒈𝒊𝒕

G. Applications of Palindromes
Palindromes in real life scenarios can have restrictions on their value.

A watch will only have certain digits used for time:

➢ if 𝑋𝑌 is the number of minutes on a clock, 𝑋 will never 6 or greater.
➢ if 𝐴𝐵 is the number of hours on a clock, 𝐴 will always be either zero or one.

Example 4.115
A palindrome is a whole number that reads the same forwards and backwards. If one neglects the colon, certain
times displayed on a digital watch are palindromes. Three examples are: 1:01, 4:44, and 12:21. How many times
during a 12-hour period will be palindromes? (AMC 8 1988/25)

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The question does not specify whether the clock is a 12-hour clock, or a 24-hour clock. However, since it says
12-hour period, the answer for all 12-hours periods should be the same, and hence, we consider the time
𝑓𝑟𝑜𝑚 00: 00 𝑡𝑜 11: 59
Case I: 𝒂𝒃𝒂
The units digit(hours) can take values from 1 to 9, giving us nine choices. This automatically decides the unit’s
digit for the minutes.
The maximum value that the minutes can take is 59. Hence, the ten’s digit can take values from 0 to 5, giving 6
choices.
We can combine the above two to get:
⏟
9 × ⏟
6 × ⏟
1 = 9 × 6 = 54
𝑼𝒏𝒊𝒕𝒔 𝑫𝒊𝒈𝒊𝒕 𝑻𝒆𝒏′ 𝒔 𝑫𝒊𝒈𝒊𝒕 𝑼𝒏𝒊𝒕′ 𝒔 𝑫𝒊𝒈𝒊𝒕
𝑯𝒐𝒖𝒓𝒔 𝑴𝒊𝒏𝒖𝒕𝒆𝒔 𝑴𝒊𝒏𝒖𝒕𝒆𝒔
Case II: 𝒂𝒃𝒃𝒂
This case has more restrictions.
The only value that the ten’s digits for the hours can take is 1, giving us one choice. Automatically, the unit’s
digit (minutes) is chosen.
The unit’s digit (hours) can take values {0,1,2} giving us three choices.
We can combine the above to get:
⏟
1 × ⏟
3 × ⏟
1 × ⏟
1 =3
𝑻𝒆𝒏𝒔 𝑫𝒊𝒈𝒊𝒕 𝑼𝒏𝒊𝒕𝒔 𝑫𝒊𝒈𝒊𝒕 𝑻𝒆𝒏′ 𝒔 𝑫𝒊𝒈𝒊𝒕 𝑼𝒏𝒊𝒕′ 𝒔 𝑫𝒊𝒈𝒊𝒕
𝑯𝒐𝒖𝒓𝒔 𝑯𝒐𝒖𝒓𝒔 𝑴𝒊𝒏𝒖𝒕𝒆𝒔 𝑴𝒊𝒏𝒖𝒕𝒆𝒔
Total = 54 + 3 = 57

In the previous question, in Case I, explain why we did not consider Unit’s Digit zero, which would give us the
following five cases like
00: 10, 00: 20, 00: 30, 00: 40, 00: 50

0010 (and the other numbers in the list above) are not palindromes.

H. Counting Digits

Example 4.116
In how many numbers does the digit 2 appear in the page numbers of the first ninety-nine pages of a book?

Method I: Enumeration

(0 − 9): 10 𝐹𝑖𝑟𝑠𝑡 𝐷𝑖𝑔𝑖𝑡𝑠 Total

10 20 . . . 90
1 11 21 91
2 12 22 32 42 . 92 10
. . . .
. . . .
9 19 29 99

1 1 10 . . . 1 19

Method II: Multiplication Principle

𝑁𝑜. 𝑜𝑓 𝐷𝑖𝑔𝑖𝑡𝑠: ⏟
10 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑓𝑜𝑟 𝑈𝑛𝑖𝑡𝑠 𝐷𝑖𝑔𝑖𝑡 + ⏟
10 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑓𝑜𝑟 𝑇𝑒𝑛𝑠 𝐷𝑖𝑔𝑖𝑡 = 10 + 10 = 20
𝟐 𝒊𝒏 𝒕𝒉𝒆 𝑻𝒆𝒏𝒔 𝑫𝒊𝒈𝒊𝒕 𝟐 𝒊𝒏 𝒕𝒉𝒆 𝑼𝒏𝒊𝒕𝒔 𝑫𝒊𝒈𝒊𝒕

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𝑁𝑜. 𝑜𝑓 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 = 20
⏟ − 1 = 19
𝑵𝒐.𝒐𝒇 𝑫𝒊𝒈𝒊𝒕𝒔

Example 4.117
How many times does the digit 2 appear in the first hundred page numbers when numbering a book?

We use the digit 2 twice in 22. So, the answer is:

19 + 1 = 20

Example 4.118
How many whole numbers between 100 and 400 contain the digit 2? (AMC 8 1985/15)

19
⏟ + 100
⏟ + 19
⏟ = 138
𝑭𝒓𝒐𝒎 𝟏−𝟏𝟎𝟎 𝑭𝒓𝒐𝒎 𝟐𝟎𝟎 𝒕𝒐 𝟐𝟗𝟗 𝑭𝒓𝒐𝒎 𝟑𝟎𝟎 𝒕𝒐 𝟑𝟗𝟗

How many whole numbers between 100 and 400 contain the digit 2? (AMC 8 1985/15)

Counting Lists
Numbers between 100 and 400 = 399 − 101 + 1 = 299 Numbers
Multiplication Principle
Nos. from 100 till 399 without 2: ⏟
2 × ⏟
9 × ⏟
9 = 2 × 9 × 9 = 162
𝑯𝒖𝒏𝒅𝒓𝒆𝒅′𝒔 𝑫𝒊𝒈𝒊𝒕 𝑻𝒆𝒏′ 𝒔 𝑫𝒊𝒈𝒊𝒕 𝑼𝒏𝒊𝒕′ 𝒔 𝑫𝒊𝒈𝒊𝒕
𝟏,𝟑
Subtract 1 from the numbers from 100 till 399 because we are not counting 100:
162 − 1 = 161
Complementary Counting
Numbers with 2 = 299 − 161 = 138

Example 4.119
Pat Peano has plenty of 0's, 1's, 3's, 4's, 5's, 6's, 7's, 8's and 9's, but he has only twenty-two 2's. How far can he
number the pages of his scrapbook with these digits? (AMC 8 1993/22)

To write the numbers from 1 − 100 will need

20 𝐷𝑖𝑔𝑖𝑡𝑠
And we write:
102
⏟ , 112
⏟
21𝑠𝑡 2 22𝑛𝑑 2
And then our 2’s are over,but we can keep writing till we need a 2, which means we can write till
119

Example 4.120
The sum of all the digits of the integers from 98 to 101 is
9 + 8 + 9 + 9 + 1 + 0 + 0 + 1 + 0 + 1 = 38
The sum of all of the digits of the integers from 1 to 2008 is (Gauss 7/2008/25)

Find the sum of the digits for 2000 − 2008:

2000 + 2001 + 2002 + ⋯ 2008 ⇒ 2(9) + (0 + 1 + 2 + ⋯ + 8) = 18 + 36 = 54

Find the sum of the digits for 0 − 1999:

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𝑇ℎ𝑜𝑢𝑠𝑎𝑛𝑑𝑠 𝐷𝑖𝑔𝑖𝑡: (0 + 1)(1000) = 1000

𝐻𝑢𝑛𝑑𝑟𝑒𝑑𝑠 𝐷𝑖𝑔𝑖𝑡: (0 + 1 + ⋯ + 9)(200) = (45)(200)
𝑇𝑒𝑛𝑠 𝐷𝑖𝑔𝑖𝑡: (0 + 1 + ⋯ + 9)(200) = (45)(200)
𝑂𝑛𝑒𝑠 𝐷𝑖𝑔𝑖𝑡: (0 + 1 + ⋯ + 9)(200) = (45)(200)

𝑇𝑜𝑡𝑎𝑙 = 54 + 1000 + (3)(45)(200) = 28054

I. Counting Rational Numbers

Challenge 4.121
Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting
numerator and denominator. For how many rational numbers between 0 and 1 will 20! be the resulting
product? (AIME 1991/5)

𝑥
We want a fraction 𝑦 in lowest form such that:
𝑥 × 𝑦 = 20! = 2𝑎 × 3𝑏 × 5𝑐 × 7𝑑 × 11𝑒 × 13 𝑓 × 17 𝑔 × 19ℎ
Hence,
2𝑎 × 3𝑏 × 5𝑐 × 7𝑑 × 11𝑒 × 13 𝑓 × 17 𝑔 × 19ℎ
𝑥=
𝑦

𝑥
Note that since 𝑦 is in lowest terms, they cannot have any common factor.
Hence, if a single 2 is in 𝑥, then 2𝑎 must be in 𝑥.

In other words, we need to divide the set {2,3,5,7,11,13,17,19} into two parts. The first part will be in 𝑥, and the
other part will be in y.

By the multiplication principle, this can be done in:

⏟
2 × ⏟ 2 × …× ⏟
2 = 28
𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
𝑓𝑜𝑟 2 𝑓𝑜𝑟 3 𝑓𝑜𝑟 19

But note that we also want:

𝑥
0< <1
𝑦
And this will be true for exactly half of the 28 choices.
Hence, the final answer is:
28
= 27 = 128
2

J. Number of Factors
An important question in number theory is the number of factors of a number. Refer to the note on Divisors in
the Number Theory to see how counting principles are applied in Number Theory.

K. Number Bases
Refer to the note on Number Bases in Number Theory to see how counting principles are applied in Number
Theory.

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L. License Plates and Codes

Example 4.122: License Plates

Example 4.123: License Plates

Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the
second from {A,I,O}, and the third from {D,M,N,T}. When Flatville needed more license plates, they added two
new letters. The new letters may both be added to one set or one letter may be added to one set and one to
another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two
letters? (AMC 8 1999/8)

𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑁𝑜. 𝑜𝑓 𝐿𝑖𝑐𝑒𝑛𝑠𝑒 𝑃𝑙𝑎𝑡𝑒𝑠 = 5 × 3 × 4 = 60

Now, to maximize the increase in the new license scheme, we need to make numbers as close as possible.
Hence, we get
{5,3,4} → {5,4,4} → {5,5,4}
Hence, the number of license plates in the new scheme
5 × 5 × 4 = 100

Additional license plates

= 100 − 60 = 40

Example 4.124: Codes

An online meeting code has the form ⏟
𝑋𝑋𝑋 − 𝑋𝑋𝑋𝑋 − 𝑋𝑋𝑋, where each 𝑋 is a non-capitalized letter from the
𝑬𝒙𝒂𝒎𝒑𝒍𝒆: 𝒚𝒒𝒌−𝒘𝒌𝒇𝒒−𝒓𝒔𝒄
English alphabet, and the hyphens are found in all codes. ⏟
𝐿𝑒𝑡𝑡𝑒𝑟𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑟𝑒𝑝𝑒𝑎𝑡𝑒𝑑. An alternate version of the
𝑹𝒆𝒑𝒆𝒕𝒊𝒕𝒊𝒐𝒏 𝑨𝒍𝒍𝒐𝒘𝒆𝒅
code has the format 𝑋𝑋𝑋𝑋𝑋, where 𝑋 can also be a digit. Find in simplest form, the ratio of the number of codes
possible in the base version and the alternate version.

Main Version: 𝑋𝑋 … 𝑋
⏟ 26 × 26 × … × 26 = 2610
=⏟
𝟏𝟎 𝑿′ 𝒔: 𝟐𝟔 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 𝒆𝒂𝒄𝒉 𝟏𝟎 𝒕𝒊𝒎𝒆𝒔
Alternate Version: 𝑋𝑋 … 𝑋
⏟ 36 × 36 × … × 36 = 365
=⏟
𝟓 𝑿′ 𝒔: 𝟑𝟔 𝑪𝒉𝒐𝒊𝒄𝒆𝒔 𝒆𝒂𝒄𝒉 𝟓 𝒕𝒊𝒎𝒆𝒔

2610 210 × 1310 210 × 1310 1310

= = 10 = 5
365 45 × 95 2 × 95 9
M. Cryptarithmetic

Challenge 4.125
A palindrome is a positive integer whose digits are the same when read forwards or backwards. For example,
2882 is a four-digit palindrome and 49194 is a five-digit palindrome. There are pairs of four-digit palindromes
whose sum is a five-digit palindrome. One such pair is 2882 and 9339. How many such pairs are there? (CEMC
Cayley 2001/24)

Let the four-digit palindromes be 𝑎𝑏𝑏𝑎 & 𝑐𝑑𝑑𝑐. Adding them should give a five-digit palindrome (say 𝑥𝑦𝑧𝑦𝑥).
So, we get the addition below:

𝑎 𝑏 𝑏 𝑎
+ 𝑐 𝑑 𝑑 𝑐

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𝑥 𝑦 𝑧 𝑦 𝑥

The maximum value of 𝑎 is 9, and the maximum value of b is also 9, and the maximum carryforward is 1. Hence,
the maximum value of
𝑎 + 𝑐 + 𝐶𝑎𝑟𝑟𝑦𝑓𝑜𝑟𝑤𝑎𝑟𝑑 = 19 ⇒ 𝑥 = 1
Hence, we get:
𝑎 𝑏 𝑏 𝑎
+ 𝑐 𝑑 𝑑 𝑐
1 𝑦 𝑧 𝑦 1

Value of 𝒂 + 𝒄
Note that from the leftmost column, the tens digit of 𝑎 + 𝑐 is 1, and from the rightmost column, the units digit of
𝑎 + 𝑐 is also 1. Hence:
𝑎 + 𝑐 = 11 ⇒ (𝑎, 𝑐) = (2,9), (3,8), (4,7), (5,6)
Note that after (5,6) the pairs will start to repeat, and hence we ignore them.

Value of 𝒃 + 𝒅

10 + 𝑦 = 𝑎 + 𝑐 + 𝐶𝑎𝑟𝑟𝑦𝑓𝑜𝑟𝑤𝑎𝑟𝑑
Substitute 𝑎 + 𝑐 = 11:
10 + 𝑦 = 11 + 𝐶𝑎𝑟𝑟𝑦𝑓𝑜𝑟𝑤𝑎𝑟𝑑
𝑦 = 1 + 𝐶𝑎𝑟𝑟𝑦𝑓𝑜𝑟𝑤𝑎𝑟𝑑

Case I: If there is no carryforward, then the only value that is works is 𝑏 = 𝑑 = 0 ⇒ 𝑧 = 0

𝑎 0 0 𝑎
+ 𝑐 0 0 𝑐
1 1 0 1 1

Case II: If there is a carryforward, then 𝑦 = 2. From the ten’s column

𝑏 + 𝑑 + 1 = 10 + 𝑦
𝑏 + 𝑑 + 1 = 12
𝑏 + 𝑑 = 11

𝑏 + 𝑑 = 11 ⇒ (𝑏, 𝑑) = (2,9), (3,8), (4,7), (5,6), (6,5)(7,4), (8,3), (9,2) ⇒ 8 𝑉𝑎𝑙𝑢𝑒𝑠

Note that this time we take 8 pairs, since the first and the pair result in different numbers.
2992 + 9229, 2222 + 9999

The total number of values from Case I and Case II:

1+8=9

By the multiplication principle the number of palindromes

= ⏟ 4 × ⏟ 9 = 36
𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
𝑓𝑜𝑟 𝑎+𝑐 𝑓𝑜𝑟 𝑏+𝑑

4.6 Preparing for Probability

A. At Least
An at least condition means that the 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 value must be at least that much.

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Example 4.126
Find the solution set of the following statements, and also write them as inequalities.
A. Reema rolled a standard six-sided die, and got a value, 𝑣, which was at least two.
B. Rudolph rolled a standard tetrahedral (four-sided die), and got a value, 𝑣, which was at least three.
C. In the coming week, I have my relatives visiting. So, I plan to watch a movie a day for 𝑚 days. I want to
watch a movie on at least three days.

Part A
The outcomes when we roll a standard six-sided dice are:
1, 2, 3, 4, 5, 6
And of the above, the outcomes which meet the conditions:
{2,3,4,5,6}

As an inequality, we will write:

2 ≤ 𝑣 ≤ 6, 𝑣∈ℕ
Part B
The outcomes when we roll a standard four-sided dice are:
1, 2, 3, 4
And of the above, the outcomes which meet the conditions:
{3,4}

As an inequality, we will write:

3 ≤ 𝑣 ≤ 4, 𝑣∈ℕ
Part C
0,1,2,3,4,5,6,7

{3,4,5,6,7}

3 ≤ 𝑥 ≤ 7, 𝑥∈ℕ
B. At Most
An at most condition means that the 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 value can be what is given

Example 4.127
Find the solution set of the following statement:
Hari draws a card from a pack of cards. He gets a card which has a value of at most 5. (Consider the Ace to have
a value of 1 for this question).

A suit of cards can be listed as below:

𝐴𝑐𝑒
⏟ , 2, 3, 4, 5, 6, 7, 8, 9, 10, 𝐽𝑎𝑐𝑘, 𝐾𝑖𝑛𝑔, 𝑄𝑢𝑒𝑒𝑛
=1

{𝐴𝑐𝑒, 2, 3, 4, 5}

C. Tossing Coins

Example 4.128
How many outcomes are there if:
A. I toss a coin

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B. I toss a coin twice

C. I toss a coin thrice

Part A
{𝐻, 𝑇} ⇒ 2 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠
Parts B
{𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇} ⇒ ⏟
2 × ⏟
2 ⇒ 4 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑
𝑇𝑜𝑠𝑠 𝑇𝑜𝑠𝑠
Parts C
{𝐻𝐻𝐻, 𝐻𝐻𝑇, 𝐻𝑇𝐻, 𝐻𝑇𝑇, 𝑇𝐻𝐻, 𝑇𝐻𝑇, 𝑇𝑇𝐻, 𝑇𝑇𝑇}
⏟
2 × ⏟ 2 × ⏟ 2 ⇒ 8 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑 𝑇ℎ𝑖𝑟𝑑
𝑇𝑜𝑠𝑠 𝑇𝑜𝑠𝑠 𝑇𝑜𝑠𝑠

Example 4.129
What is the number of outcomes if I toss a coin 𝑛 times?

This is multiplication principle with repetition. Each coin toss has 2 outcomes. Total number of outcomes is:
⏟
2 × ⏟ 2 × …× ⏟ 2 = 2𝑛 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑 𝑛𝑡ℎ
𝑇𝑜𝑠𝑠 𝑇𝑜𝑠𝑠 𝑇𝑜𝑠𝑠

Example 4.130
I toss a coin 𝑛 times, and the number of outcomes is 𝑥. Find the number of outcomes when I toss it 𝑛 + 1 times.

Method I: Direct Calculation

When we toss it 𝑛 times, the number of outcomes is:
⏟
2 × ⏟ 2 ×…× ⏟ 2 = 2𝑛 = 𝑥 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑 𝑛𝑡ℎ
𝑇𝑜𝑠𝑠 𝑇𝑜𝑠𝑠 𝑇𝑜𝑠𝑠
When we toss it 𝑛 + 1 times, the number of outcomes is:
⏟2 × ⏟ 2 × …× ⏟ 2 = 2𝑛+1 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑 (𝑛+1)𝑡ℎ
𝑇𝑜𝑠𝑠 𝑇𝑜𝑠𝑠 𝑇𝑜𝑠𝑠

2𝑛 = 𝑥
2 ∙ 2𝑛 = 2𝑥
2𝑛+1 = 2𝑥
Method I: Recursion
If you toss is 𝑛 times, the number of outcomes is 𝑥.
The (𝑛 + 1)𝑠𝑡 toss can be heads, or tails, which is two outcomes.

Each of heads or tails can be combined with the x outcomes from before. So, the total outcomes are:
2 × 𝑥 = 2𝑥

4.131: Number of Outcomes when tossing a coin: Recursive Formula

If a toss a coin 𝑛 times, the number of outcomes 𝑁 is given by
𝑁𝑛 = 2𝑁𝑛−1

Example 4.132

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I toss a coin twice. What is the number of outcomes which are:

A. No Restrictions
B. All Heads
C. All Tails
D. At least one tail
E. At most one Head
F. Exactly one tail

Counting Outcomes
Let’s write out the outcomes, using 𝐻 for Heads, and 𝑇 for Tails.
𝐵𝑜𝑡ℎ 𝐻𝑒𝑎𝑑𝑠 = 𝐻𝐻
𝐵𝑜𝑡ℎ 𝑇𝑎𝑖𝑙𝑠 = 𝑇𝑇
𝐹𝑖𝑟𝑠𝑡 𝐻𝑒𝑎𝑑, 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 𝑇𝑎𝑖𝑙𝑠 = 𝐻𝑇
𝐹𝑖𝑟𝑠𝑡 𝑇𝑎𝑖𝑙𝑠, 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 𝐻𝑒𝑎𝑑𝑠 = 𝑇𝐻

{𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇}

No Restrictions 4 𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇

All Heads 1 𝐻𝐻
All Tails 1 𝑇𝑇
≥ Condition At least one tail 3 𝐻𝑇, 𝑇𝐻, 𝑇𝑇
≤ Condition At most one Head 3 𝐻𝑇, 𝑇𝐻, 𝑇𝑇
Exactly one tail 2 𝐻𝑇, 𝑇𝐻

Multiplication Principle
⏟
2 × ⏟
2 =4
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅
𝑻𝒐𝒔𝒔 𝑻𝒐𝒔𝒔

Example 4.133
I toss the same coin three times. What is the number of possible outcomes such that:
A. There are no restrictions
B. there is at least one head in the tosses.
C. there is at least one tail in the tosses
D. there is at least one tail and one head in the tosses

Part A
Enumeration
If we toss two coins, we get the following four outcomes:
𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇
Now imagine, that the above are the second and the third coin that we toss. The first coin can be either Heads,
or Tails.

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And hence, we get the following 8 outcomes:

𝐻𝐻𝐻, 𝐻𝐻𝑇, 𝐻𝑇𝐻, 𝐻𝑇𝑇 ,
⏟ 𝑇𝐻𝐻, 𝑇𝐻𝑇, 𝑇𝑇𝐻, 𝑇𝑇𝑇
⏟
𝑭𝒊𝒓𝒔𝒕 𝑶𝒖𝒕𝒄𝒐𝒎𝒆:𝑯𝒆𝒂𝒅𝒔 𝑭𝒊𝒓𝒔𝒕 𝑶𝒖𝒕𝒄𝒐𝒎𝒆:𝑻𝒂𝒊𝒍𝒔
Multiplication Principle
⏟
2 × ⏟ 2 2 = 23 = 8
× ⏟
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝑻𝒉𝒊𝒓𝒅
𝑻𝒐𝒔𝒔 𝑻𝒐𝒔𝒔 𝑻𝒐𝒔𝒔
Part B
The outcome with all tails is not what we want:
(𝐻𝐻𝐻)(𝐻𝐻𝑇)(𝐻𝑇𝐻)(𝐻𝑇𝑇)(𝑇𝐻𝐻)(𝑇𝐻𝑇)(𝑇𝑇𝐻)(𝑻𝑻𝑻)
Use Complementary Counting:
𝑁𝑜. 𝑜𝑓 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = ⏟
8 − ⏟
1 =7
𝑻𝒐𝒕𝒂𝒍 𝑶𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝒘𝒊𝒕𝒉
𝑶𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝒂𝒍𝒍 𝑻𝒂𝒊𝒍𝒔
Part C
Complementary Counting
𝑁𝑜. 𝑜𝑓 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = ⏟
8 − ⏟
1 =7
𝑻𝒐𝒕𝒂𝒍 𝑶𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝒘𝒊𝒕𝒉
𝑶𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝒂𝒍𝒍 𝑯𝒆𝒂𝒅𝒔
Part D
Complementary Counting
𝑁𝑜. 𝑜𝑓 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = ⏟
8 − ⏟
1 − ⏟
1 =6
𝑻𝒐𝒕𝒂𝒍 𝑶𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝒘𝒊𝒕𝒉 𝑶𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝒘𝒊𝒕𝒉
𝑶𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝒂𝒍𝒍 𝑯𝒆𝒂𝒅𝒔 𝒂𝒍𝒍 𝑻𝒂𝒊𝒍𝒔

Example 4.134
I toss identical three coins in one toss.
A. What is the number of distinguishable outcomes?
B. Are these outcomes equally likely?

Part A
When you toss three identical coins in one toss, you cannot distinguish between the different arrangements.
Therefore, the only outcome that you can measure is the number of heads
➢ Zero Heads
➢ One Head
➢ Two Heads
➢ Three Heads
Giving you four outcomes in all.

Part B
Not equally likely

Example 4.135
a penny, a nickel and a dime
in that particular order
in any order that I want, and the order of the coin tosses matters

Part A-I
𝑃𝑎𝑟𝑡 𝐴: ⏟
2 × ⏟
2 2 = 23 = 8
× ⏟
𝑷𝒆𝒏𝒏𝒚 𝑵𝒊𝒄𝒌𝒆𝒍 𝑫𝒊𝒎𝒆
You can list the outcomes

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{( 𝐻
⏟ 𝐻
⏟ ⏟ ) (𝐻𝐻𝑇)(𝐻𝑇𝐻)(𝐻𝑇𝑇)(𝑇𝐻𝐻)(𝑇𝐻𝑇)(𝑇𝑇𝐻)(𝑇𝑇𝑇)}
𝐻
𝑃𝑒𝑛𝑛𝑦 𝑁𝑖𝑐𝑘𝑒𝑙 𝐷𝑖𝑚𝑒
Part A-II
In this part, I have two choices. I first need to decide the order of the coins. This I can do in six ways:
{(𝑃𝑁𝐷)(𝑃𝐷𝑁)(𝑁𝑃𝐷)(𝑁𝐷𝑃)(𝐷𝑃𝑁)(𝐷𝑁𝑃)} First Toss
Or I can also get six ways using the multiplication principle:
⏟
3 × ⏟ 2 × ⏟ 1 =6 Head Tail
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑 𝑇ℎ𝑖𝑟𝑑
𝐶𝑜𝑖𝑛 𝐶𝑜𝑖𝑛 𝐶𝑜𝑖𝑛
Head Tail Head Tail
The number of sequences for heads and tails, as calculated above is
8
Head Head Head Head
So, the total number of outcomes is
6 × 8 = 48 Tail Tail Tail Taail
D. Patterns

Example 4.136
I have a 1794 Flowing Hair Silver Dollar, a highly valuable coin among the first ones issued by the US Mint. The
obverse side has the Bust of Liberty engraved on it. The reverse side has an eagle surrounded by a wreath. I toss
the coin thrice. Count the number of outcomes with:
A. Zero Eagles
B. One Eagles
C. Two Eagles
D. Three Eagles

The language is complicated, but just think of Eagles as Heads, and Bust of Liberty as Tails.

We already know the outcomes when we toss three coins is given by:
(𝐻𝐻𝐻)(𝐻𝐻𝑇)(𝐻𝑇𝐻)(𝐻𝑇𝑇)(𝑇𝐻𝐻)(𝑇𝐻𝑇)(𝑇𝑇𝐻)(𝑇𝑇𝑇)

For this question, we need to classify the outcomes, which is done below:
𝐴: 𝑇𝑇𝑇
⏟ → 1 𝑂𝑢𝑡𝑐𝑜𝑚𝑒
𝒁𝒆𝒓𝒐
𝑯𝒆𝒂𝒅𝒔
𝐵: ⏟
𝐻𝑇𝑇, 𝑇𝐻𝑇, 𝑇𝑇𝐻 → 3 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑶𝒏𝒆 𝑯𝒆𝒂𝒅
𝐶: ⏟
𝐻𝐻𝑇, 𝐻𝑇𝐻, 𝑇𝐻𝐻 → 3 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑻𝒘𝒐 𝑯𝒆𝒂𝒅𝒔
⏟ ,
𝐻𝐻𝐻
𝑻𝒉𝒓𝒆𝒆
𝑯𝒆𝒂𝒅𝒔

Part B and Part C have the same answer above. Explain why?

Two heads mean one tail.

And two tails mean one heads.

E. Rolling a Single Die

Example 4.137: Enumerating Outcomes

I roll a standard six-sided die once. State the outcomes.

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{1, 2, 3, 4, 5, 6}

Example 4.138: Counting Outcomes

How many of the 6 outcomes when rolling a standard six-sided die once are:
A. Parity
a. Even
b. Odd
B. Prime and Composite Numbers
a. Prime
b. Composite
C. Ranges
a. Greater than four
b. Less than four
D. Ranges
a. Greater than or equal to two
b. Less than or equal to two

{1, 2, 3, 4, 5, 6}
Part A
{2,4,6} ⇒ 𝐸𝑣𝑒𝑛 = 3
{1,3,5} ⇒ 𝑂𝑑𝑑 = 3
Part B
Recall that 1 is neither prime nor composite.
{2,3,5} ⇒ 𝑃𝑟𝑖𝑚𝑒 = 3
{4,6} ⇒ 𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒 = 2
Part C
{5,6} ⇒ 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 4 = 2
{1,2,3} ⇒ 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 3 = 3
Part D
{2, 3, 4, 5, 6} ⇒ 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 2 = 5
{1,2} ⇒ 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 2 = 2

In each part (A, B, C, D, etc), what is the total? Does it add up to six each time? Why or why not?

Part B has “1” missing from the outcomes.

Part C has “4” missing from the outcomes.
Part D has “2” repeated in outcomes.

Hence, number of outcomes does not add up to 6.

F. Rolling Two Dice

Example 4.139: Outcomes

I roll two standard six-sided dice with faces numbered from one to six. Count the outcomes using:
A. Enumeration
B. Multiplication Principle

Enumeration

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Consider the outcomes in the

➢ top green row as the number from the first die.
➢ left green column as the number from the
second die. Pair of Numbers
1 2 3 4 5 6
We can combine the two to get a pair of numbers. 1 (1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
For example 2 (1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
(1,3) → 𝐹𝑖𝑟𝑠𝑡 𝑅𝑜𝑙𝑙 = 1, 𝑆𝑒𝑐𝑜𝑛𝑑 𝑅𝑜𝑙𝑙 = 3 3 (1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
Multiplication Principle 4 (1,4) (2,4) (3,4) (4,4) (5,4) (6,4)
Order is important here, since we can distinguish 5 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
between the first roll and the second roll as two 6 (1,6) (2,6) (3,6) (4,6) (5,6) (6,6)
distinct objects.
𝑇𝑜𝑡𝑎𝑙 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = ⏟
6 × ⏟
6 = 36
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅
𝑹𝒐𝒍𝒍 𝑹𝒐𝒍𝒍

G. Restrictions: Prime And Composite Rolls

Example 4.140: Prime and Composite Numbers

I have a standard six-sided die with faces numbered one to six. I roll it twice, and record the two outcomes.
What is the number of outcomes where:
A. The first roll is a prime number
B. The second roll is a composite number
C. Both rolls are prime numbers
D. Both rolls are composite numbers

𝑃𝑟𝑖𝑚𝑒𝑠 ∈ {2, 3,5} = 3 𝑃𝑟𝑖𝑚𝑒𝑠 Pair of Numbers

𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒 ∈ {4,6} = 2 𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 1 2 3 4 5 6
Part A 1 (1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
We can also do this using the multiplication rule: 2 (1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
⏟
3 × ⏟ 6 = 18 3 (1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 4 (1,4) (2,4) (3,4) (4,4) (5,4) (6,4)
𝑹𝒐𝒍𝒍 𝑹𝒐𝒍𝒍
5 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
This is also shaded in the table alongside.
6 (1,6) (2,6) (3,6) (4,6) (5,6) (6,6)
Part B
If the second roll is a composite number, we have two
options for the second roll. Pair of Numbers
And there are no restrictions on the first roll, so the total 1 2 3 4 5 6
outcomes are: 1 (1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
⏟
6 × ⏟ 2 = 12 2 (1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 3 (1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
𝑹𝒐𝒍𝒍 𝑹𝒐𝒍𝒍 (1,4) (2,4) (3,4) (4,4) (5,4) (6,4)
4
This is also shaded in the table alongside. (1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
5
6 (1,6) (2,6) (3,6) (4,6) (5,6) (6,6)

Part C
Pair of Numbers
1 2 3 4 5 6
1 (1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
2 (1,2) (2,2) (3,2) (4,2) (5,2) (6,2)

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There are three possible prime numbers 3 (1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
➢ for the first roll. 4 (1,4) (2,4) (3,4) (4,4) (5,4) (6,4)
➢ And also for the second roll 5 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
And, by the multiplication principle, the total number of 6 (1,6) (2,6) (3,6) (4,6) (5,6) (6,6)
outcomes is:
⏟
3 × ⏟ 3 =9
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅
𝑹𝒐𝒍𝒍 𝑹𝒐𝒍𝒍
This is also shaded alongside, where we want the intersection of the prime rows, and the prime columns.

Part D
⏟
2 × ⏟
2 =4 Pair of Numbers
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅
𝑹𝒐𝒍𝒍 𝑹𝒐𝒍𝒍 1 2 3 4 5 6
This is also shaded alongside, where we want the 1 (1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
intersection of the composite rows, and the composite 2 (1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
columns. 3 (1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
4 (1,4) (2,4) (3,4) (4,4) (5,4) (6,4)
5 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
6 (1,6) (2,6) (3,6) (4,6) (5,6) (6,6)

Example 4.141
I have a standard six-sided die with faces numbered one to six. I roll it twice, and record the two outcomes.
What is the number of outcomes:
A. At least one roll is a prime number
B. At most one roll is a prime number
C. Exactly one roll is a prime number

Part A: Using Casework

We can do this on the basis of casework, but we have to be careful in counting our cases.
At least one roll is a prime number has three sub-cases:
➢ Only First Roll is a prime Number = 3 × 3 = 9
➢ Only Second Roll is a Prime Number = 3 × 3 = 9
➢ Both Rolls are prime Numbers = 3 × 3 = 9

𝑇𝑜𝑡𝑎𝑙 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 9 + 9 + 9 = 27
Part A: Complementary Counting
At least one roll is a prime number can also be thought of Pair of Numbers
as 1 2 3 4 5 6
𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠(𝑁𝑜 𝑃𝑟𝑖𝑚𝑒 𝑁𝑢𝑚𝑏𝑒𝑟𝑠) = ⏟ 3 × ⏟ 3 =9 1 (1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅
𝑹𝒐𝒍𝒍 𝑹𝒐𝒍𝒍 2 (1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
36
⏟ − ⏟
9 = 27 3 (1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
𝑻𝒐𝒕𝒂𝒍 𝑵𝒐 𝑷𝒓𝒊𝒎𝒆 4 (1,4) (2,4) (3,4) (4,4) (5,4) (6,4)
𝑶𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝑵𝒖𝒎𝒃𝒆𝒓𝒔
5 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
And the numbers where neither of the rolls are prime, are
6 (1,6) (2,6) (3,6) (4,6) (5,6) (6,6)
those which fall in both a shaded column, and a shaded
row.
Part B
Again, use complementary counting:
36
⏟ − ⏟
9 = 27
𝑻𝒐𝒕𝒂𝒍 𝑩𝒐𝒕𝒉 𝑷𝒓𝒊𝒎𝒆
𝑶𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝑵𝒖𝒎𝒃𝒆𝒓𝒔

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Part C
Complementary Counting
36
⏟ − ⏟
9 − ⏟
9 = 18
𝑻𝒐𝒕𝒂𝒍 𝑵𝒐 𝑷𝒓𝒊𝒎𝒆 𝑩𝒐𝒕𝒉 𝑷𝒓𝒊𝒎𝒆
𝑶𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝑵𝒖𝒎𝒃𝒆𝒓𝒔 𝑵𝒖𝒎𝒃𝒆𝒓𝒔

Direct Counting
3×3
⏟ + 3×3
⏟ = 9 + 9 = 18
𝑭𝒊𝒓𝒔𝒕 𝑹𝒐𝒍𝒍−𝑷𝒓𝒊𝒎𝒆 𝑭𝒊𝒓𝒔𝒕 𝑹𝒐𝒍𝒍−𝑵𝒐𝒕 𝑷𝒓𝒊𝒎𝒆
𝑺𝒆𝒄𝒐𝒏𝒅 𝑹𝒐𝒍𝒍−𝑵𝒐𝒕 𝑷𝒓𝒊𝒎𝒆 𝑺𝒆𝒄𝒐𝒏𝒅 𝑹𝒐𝒍𝒍−𝑷𝒓𝒊𝒎𝒆

H. Sum

Example 4.142: Outcomes

I roll two standard six-sided dice with faces numbered from one to six. Enumerate the outcomes of the sum of
the two numbers.

Total
1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12

Example 4.143
I have a standard six-side die with faces numbered one to six. I roll it twice, and record the two outcomes. What
is the number of outcomes where the sum of the numbers is 7?

Table
We can also pick up the answer from the table. Total
See that the number seven make a diagonal, or stair-climbing B. 1 2 3 4 5 6
pattern. 1 2 3 4 5 6 7
Diophantine Equation 2 3 4 5 6 7 8
𝑎 + 𝑏 = 7, 1 ≤ 𝑎, 𝑏 ≤ 6, 𝑎, 𝑏 ∈ ℕ 3 4 5 6 7 8 9
This is a Diophantine equation, which we study separately in 4 5 6 7 8 9 10
number theory. It asks for integer solutions, with constraints, to 5 6 7 8 9 10 11
equations. 6 7 8 9 10 11 12
𝑎 ∈ {1,2,3,4,5,6}, 𝑏 ∈ {6,5,4,3,2,1} ⇒ 6 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠

Example 4.144
I have a standard six-sided die with faces numbered one to six. I roll it twice, and record the two outcomes.
What is the number of outcomes where the sum of the numbers is greater than 10?

Total
1 2 3 4 5 6 Total
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9

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Table 4 5 6 7 8 9 10
We shade the required three outcomes in the, table, 5 6 7 8 9 10 11 1
forming a triangle pattern. 6 7 8 9 10 11 12 2
Algebra
13 > 𝑎 + 𝑏 > 10, 1 ≤ 𝑎, 𝑏 ≤ 6, 𝑎, 𝑏 ∈ ℤ
This is a Diophantine Inequality
𝑎 + 𝑏 = 11 ⇒ 𝑎 ∈ {5,6}, 𝑏 ∈ {6,5}
𝑎 + 𝑏 = 12 ⇒ 𝑎 ∈ {6}, 𝑏 ∈ {6} ⇒ 3 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠

Example 4.145
I have a standard six-sided die with faces numbered one to six. I roll it twice, and record the two outcomes.
What is the number of outcomes where the sum of the numbers is less than 6?

Fixing the First Number

We can fix the first number in the die roll. Total
If the first number is 1, the second number can be: B. 1 2 3 4 5 6
1 2 3 4 5 6 7 4
2, 3, 4 𝑜𝑟 5
2 3 4 5 6 7 8 3
Giving us 4 options in all.
3 4 5 6 7 8 9 2
4 5 6 7 8 9 10 1
And we can keep counting.
5 6 7 8 9 10 11
See the table to the left. 6 7 8 9 10 11 12
10
Fixing the Total
As a different method, we can fix the total. If the total is 5, Total
we can get: B. 1 2 3 4 5 6
4+1 1 2 3 4 5 6 7 1
2+3 2 3 4 5 6 7 8 2
3+2 3 4 5 6 7 8 9 3
1+4 4 5 6 7 8 9 10 4
Giving us 4 options. 5 6 7 8 9 10 11
And we can keep counting. See the table to the left. 6 7 8 9 10 11 12

And hence the final answer is:

1 + 2 + 3 + 4 = 10
I. Parity in the Sum
Parity refers to the concept of even and odd. This is a very simple, yet very powerful concept with broad
application in Maths.

Example 4.146
I have a standard six-sided die with faces numbered one to six. I roll it twice, and record the two outcomes.
What is the number of outcomes where the sum of the numbers is odd?

Total
B. 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10

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Table 5 6 7 8 9 10 11
Shade the odd outcomes in the table, and see that we get 6 7 8 9 10 11 12
chessboard pattern.
Hence, for every odd number, there is one even number.

And hence, the number of odd numbers is:

36
= 18
2
Diophantine Inequality
This is a Diophantine Inequality, which can be solved by considering each of the cases that satisfy it.
𝑎 + 𝑏 ≤ 12, 𝑎 + 𝑏 ∈ 2𝑛 + 1, 𝑛∈𝑁

𝑎+𝑏 ={ ⏟
3 , ⏟
5 , ⏟
7 , ⏟
9 , 11
⏟ } ⇒ 2 + 4 + 6 + 4 + 2 = 18
2 𝐶𝑎𝑠𝑒𝑠 4 𝐶𝑎𝑠𝑒𝑠 6 𝑐𝑎𝑠𝑒𝑠 4 𝐶𝑎𝑠𝑒𝑠 2 𝐶𝑎𝑠𝑒𝑠

Example 4.147
I have a standard six-sided die with faces numbered one to six. I roll it twice, and record the two outcomes.
What is the number of outcomes where the sum of the numbers is even?

Enumeration

𝑎+𝑏
2 (1,1) 1
4 (1,3)(2,2)(3,1) 3 4
6 (1,5)(2,4)(3,3)(4,2)(5,1) 5 9
8 (2,6)(3,5)(4,4)(5,3)(6,2) 5 14
10 (4,6)(5,5)(6,4) 3 17
12 (6,6) 1 18

Counting Argument
𝑂𝑑𝑑 + 𝑂𝑑𝑑 = 𝐸𝑣𝑒𝑛 ⇒ 3 × 3 = 9 Total
𝐸𝑣𝑒𝑛 + 𝐸𝑣𝑒𝑛 = 𝐸𝑣𝑒𝑛 ⇒ 3 × 3 = 9 B. 1 2 3 4 5 6
1 2 3 4 5 6 7
𝑇𝑜𝑡𝑎𝑙 = 18 2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
J. Product 6 7 8 9 10 11 12

Example 4.148: Outcomes

I roll two standard six-sided dice with faces numbered from one to six. Enumerate the product of the two
numbers.

Total
B. 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36

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Example 4.149
I have a standard six-sided die with faces numbered one to six. I roll it twice, and record the two outcomes.
What is the number of outcomes where the product of the numbers is 12?

Number Theory
We can determine the number of outcomes by creating factor Total
pairs for 12. 1 2 3 4 5 6
12 = 1⏟× 12 = ⏟2×6 = ⏟ 3×4 ⇒2+2=4 1 1 2 3 4 5 6
𝑁𝑜𝑡 𝑉𝑎𝑙𝑖𝑑 𝑉𝑎𝑙𝑖𝑑 𝑉𝑎𝑙𝑖𝑑 2 2 4 6 8 10 12
(2,6),(6,2) (3,4)(4,3) 3 3 6 9 12 15 18
4 4 8 12 16 20 24
(𝟐, 𝟔), (𝟔, 𝟐) 5 5 10 15 20 25 30
6 6 12 18 24 30 36
Note that out of the factor pairs, the first factor pair is not valid,
since 12 is not a valid outcome when rolling a standard six-sided die.

Example 4.150
I have a standard six-sided die with faces numbered one to six. I roll it twice, and record the two outcomes.
A. What is the number of outcomes where the product of the two rolls is greater than 25?
B. If the product of the two rolls is greater than 25, than the sum of the two rolls must be greater than or
equal to which number?

Part A
Product should be greater than 25: Total
(5,6)(6,5)(6,6) ⇒ 3 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
Part B
Sum must be
Total
≥ 11
1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12

K. Parity in the Product

𝑂𝑑𝑑 × 𝑂𝑑𝑑 = 𝑂𝑑𝑑 𝑂𝑑𝑑 × 𝐸𝑣𝑒𝑛 = 𝐸𝑣𝑒𝑛

𝐸𝑣𝑒𝑛 × 𝑂𝑑𝑑 = 𝐸𝑣𝑒𝑛
𝐸𝑣𝑒𝑛 × 𝐸𝑣𝑒𝑛 = 𝐸𝑣𝑒𝑛

Example 4.151

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I have a standard six-sided die with faces numbered one to six. I roll it twice, and record the two outcomes.
What is the number of outcomes where the product of the numbers is odd?

Multiplication Principle Total

The product of two numbers is odd only when both the numbers 1 2 3 4 5 6
are odd. 1 1 2 3 4 5 6
Hence, the number of odd products is: 2 2 4 6 8 10 12
⏟
3 × ⏟ 3 =9 3 3 6 9 12 15 18
𝑶𝒅𝒅 𝑶𝒅𝒅 4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36

Example 4.152
I have a standard six-sided die with faces numbered one to six. I roll it twice, and record the two outcomes.
What is the number of outcomes where the product of the numbers is even?

Direct Method
𝑂𝑑𝑑 × 𝐸𝑣𝑒𝑛 = 𝐸𝑣𝑒𝑛 ⇒ ⏟
3 × ⏟
3 =9 Total
𝑂𝑑𝑑 𝐸𝑣𝑒𝑛 1 2 3 4 5 6
𝐸𝑣𝑒𝑛 × 𝑂𝑑𝑑 = 𝐸𝑣𝑒𝑛 ⇒ ⏟
3 × ⏟
3 =9 1 1 2 3 4 5 6
𝐸𝑣𝑒𝑛 𝑂𝑑𝑑 2 2 4 6 8 10 12
𝐸𝑣𝑒𝑛 × 𝐸𝑣𝑒𝑛 = 𝐸𝑣𝑒𝑛 ⇒ ⏟
3 × ⏟
3 =9 3 3 6 9 12 15 18
𝐸𝑣𝑒𝑛 𝐸𝑣𝑒𝑛 4 4 8 12 16 20 24
5 5 10 15 20 25 30
Complementary Counting 6 6 12 18 24 30 36
36
⏟ − ⏟
9 = 27
𝑻𝒐𝒕𝒂𝒍 𝑬𝒗𝒆𝒏
𝑶𝒖𝒕𝒄𝒐𝒎𝒆𝒔

Example 4.153
If you a toss a standard five-sided die twice and find the product, what is the number of even outcomes?

If you see the table you realize that: Total

𝑁𝑜. 𝑜𝑓 𝑂𝑑𝑑 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 9 1 2 3 4 5
1 1 2 3 4 5
𝑇𝑜𝑡𝑎𝑙 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 25
2 2 4 6 8 10
3 3 6 9 12 15
𝐸𝑣𝑒𝑛 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 25 − 9 = 16
4 4 8 12 16 20
5 5 10 15 20 25
L. Rolling Three Dice

Example 4.154
I roll three six-sided dice (one colored red, the second colored green, and the third colored blue). What is the
number of outcomes
A. With no restrictions
B. With all prime numbers
C. With all composite numbers
D. Where the product of the three outcomes is odd
E. Where the product of the three outcomes is even
F. Where all the rolls are six

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Part A
⏟
6 × ⏟
6 × ⏟
6 = 63 = 216
𝑹𝒆𝒅 𝑫𝒊𝒆 𝑮𝒓𝒆𝒆𝒏 𝑫𝒊𝒆 𝑩𝒍𝒖𝒆 𝑫𝒊𝒆
Part B
𝑃𝑟𝑖𝑚𝑒𝑠 ∈ {2,3,5}
⏟
3 × ⏟
3 × ⏟ 3 = 33 = 27
𝑹𝒆𝒅 𝑫𝒊𝒆 𝑮𝒓𝒆𝒆𝒏 𝑫𝒊𝒆 𝑩𝒍𝒖𝒆 𝑫𝒊𝒆
Part C
𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 {4,6}
⏟
2 × ⏟
2 × ⏟2 = 23 = 8
𝑹𝒆𝒅 𝑫𝒊𝒆 𝑮𝒓𝒆𝒆𝒏 𝑫𝒊𝒆 𝑩𝒍𝒖𝒆 𝑫𝒊𝒆
Part D
𝑂𝑑𝑑 × 𝑂𝑑𝑑 × 𝑂𝑑𝑑 = 𝑂𝑑𝑑
⏟
3 × ⏟
3 × ⏟ 3 = 33 = 27
𝑹𝒆𝒅 𝑫𝒊𝒆 𝑮𝒓𝒆𝒆𝒏 𝑫𝒊𝒆 𝑩𝒍𝒖𝒆 𝑫𝒊𝒆
Part E
We will use complementary counting and subtract the number of odd outcomes (𝑃𝑎𝑟𝑡 𝐷) from the total number
of outcomes (𝑃𝑎𝑟𝑡 𝐴)
𝑇𝑜𝑡𝑎𝑙 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 − 𝑂𝑑𝑑 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 216 − 27 = 189
Part F
⏟
1 × ⏟
1 × ⏟1 = 13 = 1
𝑹𝒆𝒅 𝑫𝒊𝒆 𝑮𝒓𝒆𝒆𝒏 𝑫𝒊𝒆 𝑩𝒍𝒖𝒆 𝑫𝒊𝒆

M. Rolling Sides Other Than Six

Example 4.155
I roll two dice, one with four sides numbered one through four, and the other with eight side numbered one
through eight. What is the number of outcomes:
A. With no restrictions
B. With all prime numbers
C. With all composite numbers
D. Where the product of the two rolls is odd
E. Where the product of the two rolls is even
F. Where all the rolls are six

Part A
4 × 8 = 32
Part B
𝐹𝑜𝑢𝑟 𝑆𝑖𝑑𝑒𝑑 𝐷𝑖𝑒: {2,3}, 𝐸𝑖𝑔ℎ𝑡 𝑆𝑖𝑑𝑒𝑑 𝐷𝑖𝑒: {2,3,5,7} ⇒ 2 × 4 = 8
Part C
𝐹𝑜𝑢𝑟 𝑆𝑖𝑑𝑒𝑑 𝐷𝑖𝑒: {4}, 𝐸𝑖𝑔ℎ𝑡 𝑆𝑖𝑑𝑒𝑑 𝐷𝑖𝑒: {4,6,8} ⇒ 1 × 3 = 3
Part D
The only way in which the product will be odd is if both the dice have odd outcomes
𝑂𝑑𝑑 × 𝑂𝑑𝑑 = 𝑂𝑑𝑑
𝐹𝑜𝑢𝑟 𝑆𝑖𝑑𝑒𝑑 𝐷𝑖𝑒: {1,3}, 𝐸𝑖𝑔ℎ𝑡 𝑆𝑖𝑑𝑒𝑑 𝐷𝑖𝑒: {1,3,5,7} ⇒ 2 × 4 = 8
Part E
𝑇𝑜𝑡𝑎𝑙 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 − 𝑂𝑑𝑑 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 32 − 8 = 24
Part F
You cannot get a roll of six with a four-sided die.
Hence, zero outcomes.

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N. Standard Pack of Cards

A standard pack of 52 cards has:
➢ Four suits (Clubs, Spades, Hearts, and Diamonds), with 13 cards in each suit.
✓ Clubs and Spades are “Black” suits
✓ “Hearts” and “Diamonds” are “Red” suits.
➢ 13 ranks in each suit
✓ Ranks are: Ace, 2, 3, …8, 9, 10, Jack, Queen and King
✓ The last three are called Face Cards.
➢ 4 cards of each rank (one of each suit)

Example 4.156
Enumerate the outcomes of drawing a single card.

Each cell in the table represents one outcome:

Ace 2 3 4 5 6 7 8 9 10 Jack Queen King Total

Clubs Face Face Face 13
Card Card Card
Black
Spades Face Face Face 13
Card Card Card
Hearts Face Face Face 13
Card Card Card
Red
Diamonds Face Face Face 13
Card Card Card
4 4 4 4 4 4 4 4 4 4 4 4 4 52

Example 4.157
I draw a single card from a pack of cards. Consider Ace as 1, and face cards as not having a number. What is the
number of outcomes where the card:

A. Has no restrictions K. Is Red, but not a diamond

B. is red L. Is Black, but not diamonds or hearts
C. Is black M. Is either hearts or a face card
D. Is spades N. Is either black or not a face card
E. Is Hearts O. Is a red face card
F. is a face card P. Is a black non-face card
G. Is a non-face card Q. Is a face card that is Spades
H. Is odd numbered R. Is a non-face card that is not Spades
I. Is prime numbered S. Is a black odd numbered card
J. Is composite numbered T. Is an odd numbered Spades

Has no restrictions 52
Red 26
Black 26
Spades 13
Hearts 13
Face card 12 4×3
Not a face card 40 52 − 12
Odd Numbered 20 5×4 {1,3,5,7,9}

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Prime Numbered 16 4×4 {2,3,5,7}

Composite Numbered 20 5×4 {4,6,8,9,10}
Red, but not a Diamond 13 𝐻𝑒𝑎𝑟𝑡𝑠
Black, but not a Diamond or a 26
Heart
Either hearts or a face card 22 13 + 9 13 + 12 − 3
Either black or not a face card 46 26 + 20 52 − 6
Red Face Card 6
Black non-face card 20
Spades face card 3
Non-face Non-Spades Card 30
Black Odd Numbered Card 10
Odd Numbered Spades 5

Is either 𝑏𝑙𝑎𝑐𝑘
⏟ or ⏟
𝑛𝑜𝑡 𝑎 𝑓𝑎𝑐𝑒 𝑐𝑎𝑟𝑑
𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝐼 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝐼𝐼
When you have OR, either of the conditions being met is sufficient for the card to be included in the counting.

𝐵𝑙𝑎𝑐𝑘 = 26
The number of non-face cards:
𝑁𝑜𝑡 𝑎 𝐹𝑎𝑐𝑒 𝐶𝑎𝑟𝑑 = 4(13 − 3) = 4 × 10 = 40 ⇒ 20 𝐵𝑙𝑎𝑐𝑘, 20 𝑅𝑒𝑑

Final Answer
26
⏟ + 20
⏟ = 46
𝐵𝑙𝑎𝑐𝑘 𝑁𝑜𝑛−𝐵𝑙𝑎𝑐𝑘,𝑁𝑜𝑛−𝐹𝑎𝑐𝑒 𝐶𝑎𝑟𝑑𝑠

O. Drawing Two Cards

Example 4.158
I draw two cards, without replacement, from a standard pack of cards. What is the number of outcomes?

52
⏟ × 51
⏟ = 2,652
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑
𝐶𝑎𝑟𝑑 𝐶𝑎𝑟𝑑

Repeat the above question if the cards are drawn with replacement?
52
⏟ × 52 ⏟ = 2704
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑
𝐶𝑎𝑟𝑑 𝐶𝑎𝑟𝑑

Example 4.159
I draw two cards, with replacement, from a standard pack of cards. What is the number of outcomes where both
the cards are:
A. Red
B. Spades
C. Face cards

26 × 26 = 676
13 × 13 = 169
12 × 12 = 144

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Repeat the above question if the cards are drawn without replacement.

26
⏟ × 25
⏟
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑
𝐶𝑎𝑟𝑑 𝐶𝑎𝑟𝑑
13
⏟ × 12
⏟
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑
𝐶𝑎𝑟𝑑 𝐶𝑎𝑟𝑑
12
⏟ × 11
⏟
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑
𝐶𝑎𝑟𝑑 𝐶𝑎𝑟𝑑

Example 4.160
I draw two cards, with replacement, from a standard pack of cards. What is the number of outcomes where both
the cards are
D. either Clubs or an Ace

16 × 16 = 256

Repeat the above question if the cards are drawn without replacement.
16
⏟ × 15 ⏟ = 240
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑
𝐶𝑎𝑟𝑑 𝐶𝑎𝑟𝑑

(Important) Example 4.161

I draw two cards, without replacement, from a standard pack of cards. What is the number of outcomes where
exactly one card is:
A. A club
B. Red
C. A face card

Part A: 𝑪𝒍𝒖𝒃 = 𝑪, 𝑵𝒐𝒏 − 𝑪𝒍𝒖𝒃 = 𝑵

𝐹𝑖𝑟𝑠𝑡 𝐶𝑎𝑟𝑑 𝑖𝑠 𝑎 𝐶𝑙𝑢𝑏 = 13
⏟ × 39
⏟
𝐶 𝑁
𝑆𝑒𝑐𝑜𝑛𝑑 𝐶𝑎𝑟𝑑 𝑖𝑠 𝑎 𝐶𝑙𝑢𝑏 = 39
⏟ × 13
⏟
𝑁 𝐶
𝑇𝑜𝑡𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 = 2 × 13 × 39

Part B: 𝑹𝒆𝒅 = 𝐑, 𝑵𝒐𝒏 − 𝐑𝐞𝐝 = 𝑹

𝐹𝑖𝑟𝑠𝑡 𝐶𝑎𝑟𝑑 𝑖𝑠 𝑅𝑒𝑑 = 26
⏟ × 26
⏟ = 676
𝑅 𝑁
𝑆𝑒𝑐𝑜𝑛𝑑 𝐶𝑎𝑟𝑑 𝑖𝑠 𝑅𝑒𝑑 = 26
⏟ × 26
⏟ = 676
𝑁 𝑅
𝑇𝑜𝑡𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 = 2 × 676 = 1352

Part C: 𝑭𝒂𝒄𝒆 = 𝐅, 𝑵𝒐𝒏 − 𝐅𝐚𝐜𝐞 = 𝐍

𝐹𝑖𝑟𝑠𝑡 𝐶𝑎𝑟𝑑 𝑖𝑠 𝑎 𝐹𝑎𝑐𝑒 𝐶𝑎𝑟𝑑 = 12
⏟ × 40
⏟
𝐹 𝑁
𝑆𝑒𝑐𝑜𝑛𝑑 𝐶𝑎𝑟𝑑 𝑖𝑠 𝑎 𝐹𝑎𝑐𝑒 𝐶𝑎𝑟𝑑 = 40
⏟ × 12
⏟
𝑁 𝐶
𝑇𝑜𝑡𝑎𝑙 = 2 × 12 × 40 = 960

Repeat the above question if the cards are drawn with replacement.

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The answers here are the same as the answers without replacement.
Because you are drawing from two different categories for each question:
➢ Club versus Non-Club
➢ Red Card versus Non-Red Card
➢ Face card versus Non-Face Card

Example 4.162
I draw two cards, without replacement, from a standard pack of cards. What is the number of outcomes where
the cards are:
A. At most one is hearts
B. At least one is spades

This is best done with complementary counting.

At Most One is Hearts

(52 × 51) − ⏟
⏟ (13 × 12) = 2652 − 156 = 2496
𝐴𝑙𝑙 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝐵𝑜𝑡ℎ 𝐻𝑒𝑎𝑟𝑡𝑠
At Least One is Spades
(52 × 51) − ⏟
⏟ (39 × 38) = 2652 − 1482 = 1170
𝐴𝑙𝑙 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑁𝑜 𝑆𝑝𝑎𝑑𝑒𝑠

Answer the above questions if the cards are drawn with replacement

52 × 52 − 13 × 13 = 522 − 132 = 2535

52 × 52 − 39 × 39 = 522 − 392 = 1183

(Casework) Example 4.163

I draw two cards from a standard pack of cards, without replacement. What is the number of outcomes when
the first card is red, the second card is a face card?

26 × 12 = 312
This is not correct because the 26 red cards also include 6 face cards, and the red card that you pick could be a
face card, in which case, the number of face cards available to pick is only 11.

We have to solve this question using casework:

Case I: First Card is a Red Face Card:

6 × 11 = 66

Case II: First is a Red Non-Face Card:

26
⏟− ⏟
6 = 20
𝑅𝑒𝑑 𝑅𝑒𝑑 𝐹𝑎𝑐𝑒 𝐶𝑎𝑟𝑑𝑠
20 × 12 = 240

𝑇𝑜𝑡𝑎𝑙 = 66 + 240 = 306

P. Drawing Three Cards

Example 4.164
I draw three cards from a standard pack of cards. What is the number of outcomes when the cards are drawn:

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A. With replacement
B. Without replacement

52 × 52 × 52 = 523 = 140,608

Example 4.165
I draw three cards from a standard pack of cards. What is the number of outcomes when the first card is
Diamonds, the second card is Hearts, and the third card is Spades?

13 × 13 × 13 = 133 = 2,197

Example 4.166
I draw three cards from a standard pack of cards. What is the number of outcomes where one card is Diamonds,
one card is Hearts, and one card is Spades?

13 × 13 × 13 = 133
I can also arrange the cards among themselves.

3 × 2 × 1 = 6 𝑊𝑎𝑦𝑠
𝑇𝑜𝑡𝑎𝑙 = 6 × 133

Example 4.167
I draw three cards from a standard pack of cards, without replacement. What is the number of outcomes when
the first card is red, the second card is black, and there are no restrictions on the third card?

26 × 26 × 50 = 33,800

(Casework) Example 4.168

I draw three cards from a standard pack of cards, without replacement. What is the number of outcomes when
the first card is red, the second card is black, and the third card is a face card?

Q. Review

4.169: Total Outcomes

When we toss a coin, roll dice, draw cards, draw a ball from an urn, we want to know what the possible
outcomes.
These will later on in probability be called total outcomes (sample space)

Example 4.170
What is the number of outcomes when we toss:
A. 3 coins
B. 5 coins
C. 𝑛 coins

2 × 2 × 2 = 23 = 8
25 = 32

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2 × 2 × 2 × … .× 2 = 2𝑛
⏟
𝑛 𝑡𝑖𝑚𝑒𝑠

Example 4.171
What is the number of outcomes when we have six-sided dice and we roll:
A. Two Dice
B. Three Dice
C. Four Dice
D. 𝑛 dice

6 × 6 = 36
6 × 6 × 6 = 63 = 216
6 × 6 × 6 × 6 = 1296
6𝑛

Example 4.172
What is the number of outcomes when we roll two:
A. Octahedral (eight-sided) dice
B. Tetrahedral (four-sided) dice
C. Ten-sided dice
D. Twenty-sided dice

8 × 8 = 64
4 × 4 = 16
10 × 10 = 100
20 × 20 = 400

Example 4.173
What is the number of outcomes when we roll:
A. Three eight-sided dice
B. Four ten-sided dice
C. Two twenty-sided dice
D. Two hundred-sided dice

8 × 8 × 8 = 512
104 = 10,000
20 × 20 = 400
1002 = 10,000

Example 4.174
What is the number of outcomes when I draw, with replacement, from a standard pack of cards:
A. Two Cards
B. Three Cards
C. 𝑛 cards

522
523
52𝑛

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Example 4.175
I toss six six-sided dice. What is the number of outcomes in which:
A. Every roll is odd
B. Every roll is even
C. Every roll is prime
D. Every roll is composite
E. The rolls are in increasing order
F. The rolls are in decreasing order

36 = 729
36 = 729
36 = 729
26 = 64
1
1

Example 4.176
I toss six six-sided dice. What is the number of outcomes in which every roll is a different number?

There are six dice, and, in some order, they must be the numbers:
1,2,3,4,5,6

The first die can be any number.

6 𝑐ℎ𝑜𝑖𝑐𝑒𝑠
The second die cannot repeat the number from the first die
5 𝑐ℎ𝑜𝑖𝑐𝑒𝑠
.
.
.
The last die has only one number:
1 𝐶ℎ𝑜𝑖𝑐𝑒

Hence, we bring the choices together:

6 × 5 × … × 1 = 6! = 720

Example 4.177
I toss six six-sided dice. What is the number of outcomes in which the product of the rolls is a prime?

𝑃𝑟𝑖𝑚𝑒𝑠 ∈ {2,3,5}

Suppose we multiply a prime with any number other than one, then the product will not be a prime.
Hence, exactly one of the rolls must be prime, and all other rolls must be 1.

Consider that the prime number that we choose is 2:

(1,1,1,1,1,2), (1,1,1,1,2,1), (1,1,1,2,1,1), (1,1,2,1,1,1), (1,2,1,1,1,1), (2,1,1,1,1,1) ⇒ 6 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠

Similarly, we have:

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𝑃𝑟𝑜𝑑𝑢𝑐𝑡 = 3 ⇒ 6 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 = 5 ⇒ 6 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠

And hence the final answer is:

6 + 6 + 6 = 18 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠

Example 4.178
What is the number of outcomes when I draw, with replacement, from a standard pack of cards:
A. Two Red Cards
B. Two Spades
C. Two Face Cards
D. Two Non-Face Cards
E. Two Kings

262 = 676
132 = 169
122 = 144
402 = 1600
42 = 16

4.7 Further Applications of Counting

A. Number of Rays
You can use the rules of counting to count geometrical objects.
Recall from geometry that a ray has a single endpoint and extends towards infinity in the other direction.

Example 4.179: Collinear Points

Points 𝐴𝐵𝐶𝐷 lie on a line, in that order.
A. Which are the rays that can be formed using these points only?
B. Count the number of rays that can be formed.

For the sake of simplicity assume that 𝐴, 𝐵, 𝐶 and 𝐷 are oriented left to right.
We can then draw a diagram like the one below:

A B C D
What doesn’t work A
The most important thing is to realize that
𝑅𝑎𝑦 𝐴𝐵 = 𝑅𝑎𝑦 𝐴𝐶
Because both rays have start point A, and both rays go till infinity in the direction of B (which is the same as the
direction of C).
Hence, you cannot count the above as two different rays.

Strategy
Rays starting from A:
𝐴𝐵
Rays starting from B:
𝐵𝐴, 𝐵𝐶
Rays starting from C:

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𝐶𝐵, 𝐶𝐷
Rays starting from D:
𝐷𝐶

𝑇𝑜𝑡𝑎𝑙 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 = 1 + 2 + 2 + 1 = 6
4.180: Number of Rays on 𝒏 collinear points
If 𝒑𝟏 , 𝒑𝟐 , … 𝒑𝒏 are 𝒏 collinear points, the number of rays that can be formed using these points is:
𝟐(𝒏 − 𝟏)

If the points are not oriented left to right, rotate them to make them left to right.

Start creating rays from left to right, getting

𝑝1 𝑝2 , 𝑝2 𝑝3 , … , 𝑝𝑛−1 𝑝𝑛 ⇒ 𝑛 − 1 𝑟𝑎𝑦𝑠
The last point cannot make a ray since we only have a single point.
(We can make a single ray from each of the points expect the last one, giving us 𝑛 − 1 rays).

Repeat the process from right to left, again getting

𝑛 − 1 𝑟𝑎𝑦𝑠

The total number of rays is:

2(𝑛 − 1)

Example 4.181: Collinear Points

What is the number of rays that can be formed from 10 collinear points if at least two of the given points must
lie on the ray?

2(𝑛 − 1) = 2(10 − 1) = 2 × 9 = 18

Example 4.182: Points on a Circle

Points 𝐴𝐵𝐶 lie on a circle. Find the number of rays that can be drawn using these points if at least two of the
given points must lie on the ray?

The rays that can be made are the pairs of points:

𝐴𝐵, 𝐴𝐶, 𝐵𝐶, 𝐵𝐴, 𝐶𝐴, 𝐶𝐵 ⇒ 6 𝑃𝑎𝑖𝑟𝑠
Note that order matters here since
𝑅𝑎𝑦 𝐴𝐵 ≠ 𝑅𝑎𝑦 𝐵𝐴
Since 𝐵𝐴 goes in the exact opposite direction as 𝐴𝐵.

4.183: Number of Rays on a circle

𝒏 points lying on a circle can be used to form
𝒏(𝒏 − 𝟏) 𝑹𝒂𝒚𝒔

Each of the 𝑛 points can be made into an endpoint, and it can be connected to the remaining
𝑛 − 1 𝑝𝑜𝑖𝑛𝑡𝑠

Using the multiplication principle, we get:

⏟
𝑛 × (𝑛
⏟ − 1) 𝑅𝑎𝑦𝑠
𝑬𝒏𝒅𝒑𝒐𝒊𝒏𝒕𝒔 𝑷𝒐𝒊𝒏𝒕𝒔 𝒕𝒐
𝒄𝒐𝒏𝒏𝒆𝒄𝒕

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Example 4.184: Points on a Circle

A circle has eight distinct points spaced equally around it. Using these eight points, find the number of rays
A. Rays that can be drawn
B. Line Segments that can be drawn

Part A
⏟
8 × ⏟
7 = 56
𝑬𝒏𝒅𝒑𝒐𝒊𝒏𝒕𝒔 𝑷𝒐𝒊𝒏𝒕𝒔 𝒕𝒐
𝑪𝒐𝒏𝒏𝒆𝒄𝒕
Part B
The number of line segments will be exactly half the number of rays, since in line segments direction will not
matter, whereas it will matter for rays.

Hence, we can make pairs of rays that correspond to the same line segment

(𝑅𝑎𝑦 𝐴𝐵, 𝑅𝑎𝑦 𝐵𝐴) ⇔ 𝐿𝑖𝑛𝑒 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐴𝐵

Hence, the final answer is
56
= 28
2
B. Geometry

Example 4.185
Square 𝑃𝑄𝑅𝑆 has sides of length 8. It is split into four rectangular regions by two line segments, one parallel to
𝑃𝑄 and another parallel to 𝑄𝑅. There are 𝑁 ways in which these lines can be drawn so that the area of each of
the four rectangular regions is a positive integer. What is the remainder when 𝑁 2 is divided by 100? (Gauss
Grade 8 2021/25)

of 8:
𝑏 ∈ {1,2,4,8}
Case I: 𝒃 = 𝟏
𝑥 ∈ {1,2, … ,7} = 7 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
𝑦 ∈ {1,2, … ,7} = 7 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
𝑇𝑜𝑡𝑎𝑙 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 = 7 × 7 = 49 𝑊𝑎𝑦𝑠
(Note that 𝑦 has not been taken to have fractional
values. This will be considered in the cases below).
Case II: 𝒃 = 𝟐
1 3 15
Consider values of 𝑥 where 𝑥 is written as a fraction 𝑥 ∈ { , , … , } = 8 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
2 2 2
in reduced form: 𝑦 ∈ {2,4,6} = 3 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
𝑎
𝑥= , 𝑎, 𝑏 ∈ ℕ, 𝐻𝐶𝐹(𝑎, 𝑏) = 1 𝑇𝑜𝑡𝑎𝑙 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 = 8 × 3 = 24 𝑊𝑎𝑦𝑠
𝑏 1 3 15
Note that 𝑦 ∈ { , , … , } = 8 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
𝐴(𝑃𝑄𝑈𝑇) = 𝐴(𝑃𝑉𝑍𝑇) + 𝐴(𝑉𝑄𝑈𝑍) = 8𝑥 ∈ ℕ 2 2 2
𝑥 ∈ {2,4,6} = 3 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
Hence, we can ignore all possibilities where
𝑇𝑜𝑡𝑎𝑙 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 = 8 × 3 = 24 𝑊𝑎𝑦𝑠
𝐴(𝑃𝑄𝑈𝑇) is not a natural number.
24 + 24 = 48 𝑊𝑎𝑦𝑠
Case III: 𝒃 = 𝟒
Let: 1 3 31
𝑎 𝑥 ∈ { , , … , } = 16 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
𝐴(𝑃𝑄𝑈𝑇) = 8𝑥 = 8 ( ) 4 4 4
𝑏
𝑎 𝑦 ∈ {4} = 1 𝐶ℎ𝑜𝑖𝑐𝑒
For 8 ( ) to be a natural number, 𝑏 must be a factor
𝑏 𝑇𝑜𝑡𝑎𝑙 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 = 16 × 1 = 16 𝑊𝑎𝑦𝑠

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1 3 31 1 3
𝑦 ∈ { , , … , } = 16 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑥 ∈ { , ,…}
4 4 4 8 8
𝑥 ∈ {4} = 1 𝐶ℎ𝑜𝑖𝑐𝑒 𝑦 = 8 ⇒ 𝑁𝑜𝑡 𝑉𝑎𝑙𝑖𝑑
𝑇𝑜𝑡𝑎𝑙 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 = 16 × 1 = 16 𝑊𝑎𝑦𝑠 Hence:
16 + 16 = 32 𝑊𝑎𝑦𝑠 𝑁 = 49 + 48 + 32 = 129
Case IV: 𝒃 = 𝟖 𝑁 2 = 1292 = 16,641
𝐿𝑎𝑠𝑡 𝑡𝑤𝑜 𝑑𝑖𝑔𝑖𝑡𝑠 = 41

C. Coordinate Geometry: Lattice Points

We now turn our attention to counting in co-ordinate geometry, of which you should have at least a basic
understanding before you start this section.
Don’t memorize formulas in this section. You will most likely forget. Rather, you should be able to visualize the
question.

4.186: Lattice Points

In a cartesian co-ordinate system, the location of a point P is given by:
𝑃 = (𝑥, 𝑦)
If both 𝑥 and 𝑦 are integers, then P is called a lattice point.

Example 4.187: First Quadrant

Let 𝐴 = (2,3), 𝐵 = (7,3). Let 𝑝 be the number of lattice points between A and 𝐵, and 𝑞 be the number of lattice
points from A to B. Find 2𝑝 + 3𝑞.

𝑝= ⏟
6 − ⏟
3 + 1 = 4, 𝑞= ⏟
7 − ⏟
2 + 1 = 6 ⇒ 2𝑝 + 3𝑞 = 2(4) + 6(3) = 26
𝑺𝒕𝒂𝒓𝒕 𝑬𝒏𝒅 𝑺𝒕𝒂𝒓𝒕 𝑬𝒏𝒅
𝑷𝒐𝒊𝒏𝒕 𝑷𝒐𝒊𝒏𝒕 𝑷𝒐𝒊𝒏𝒕 𝑷𝒐𝒊𝒏𝒕

Example 4.188: Across Multiple Quadrants

Let 𝐴 = (5, −2) and 𝐵 = (5,12). Let 𝑝 be the number of lattice points between A and 𝐵, and 𝑞 be the number of
lattice points from A to B. Find −𝑝𝑞.

⏟ − (−1)
𝑝 = 11 ⏟ + 1 = 13, ⏟ − (−2)
𝑞 = 12 ⏟ + 1 = 6 ⇒ 2𝑝 + 3𝑞 = 2(4) + 6(3) = 26
𝑺𝒕𝒂𝒓𝒕 𝑬𝒏𝒅 𝑺𝒕𝒂𝒓𝒕 𝑬𝒏𝒅
𝑷𝒐𝒊𝒏𝒕 𝑷𝒐𝒊𝒏𝒕 𝑷𝒐𝒊𝒏𝒕 𝑷𝒐𝒊𝒏𝒕

4.189: Lattice Points along a Diagonal

𝑵𝒐. 𝒐𝒇 𝐋𝐚𝐭𝐭𝐢𝐜𝐞 𝑷𝒐𝒊𝒏𝒕𝒔 𝒐𝒏 𝒕𝒉𝒆 𝑫𝒊𝒂𝒈𝒐𝒏𝒂𝒍 = 𝑵𝒐. 𝒐𝒇 𝑷𝒐𝒊𝒏𝒕𝒔 𝐟𝐫𝐨𝐦 𝐨𝐧𝐞 𝐜𝐨𝐫𝐧𝐞𝐫 𝐭𝐨 𝐚𝐧𝐨𝐭𝐡𝐞𝐫 = 𝟓
See the square alongside with a side length of four, and convince yourself, using the bijection
principle that:

Example 4.190
Let 𝐴 = (5, −2) and 𝐵 = (5,12), 𝐶 = (19,12) and 𝐷 = (19, −2). Find the number of lattice points on the line
between A and C.

𝐴𝐵 = 𝐵𝐶 = 𝐶𝐷 = 𝐷𝐴 = 13 ⇒ 𝐴𝐵𝐶𝐷 𝑖𝑠 𝑎 𝑠𝑞𝑢𝑎𝑟𝑒
∴ 𝑁𝑜. 𝑜𝑓 𝐿𝑎𝑡𝑡𝑖𝑐𝑒 𝑃𝑜𝑖𝑛𝑡𝑠 = 𝑁𝑜. 𝑜𝑓 𝑃𝑜𝑖𝑛𝑡𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐴 𝑎𝑛𝑑 𝐶 = 13

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Example 4.191
(2,3)
⏟ , (6,3)
⏟ , (7,3)
⏟ , (6,7)
⏟ are the coordinates of a rectangular region in the coordinate plane
𝑩𝒐𝒕𝒕𝒐𝒎 𝑩𝒐𝒕𝒕𝒐𝒎 𝑻𝒐𝒑 𝑻𝒐𝒑
𝑳𝒆𝒇𝒕 𝑹𝒊𝒈𝒉𝒕 𝑳𝒆𝒇𝒕 𝑹𝒊𝒈𝒉𝒕
(see diagram). Find the number of lattice points on the boundaries of the region.

This method should remind you of the one used to calculate the area of a pathway in a
rectangular region. It uses the same idea, carried over to lattice points:
𝐵𝑜𝑢𝑛𝑑𝑎𝑟𝑖𝑒𝑠 = 𝑂𝑛 𝑜𝑟 𝐼𝑛𝑠𝑖𝑑𝑒
⏟ − 𝐼𝑛𝑠𝑖𝑑𝑒
⏟ = 30 − 12 = 18
𝐵𝑟𝑜𝑤𝑛+𝐺𝑟𝑒𝑒𝑛=5×6=30 𝐺𝑟𝑒𝑒𝑛=3×4
Direct Method
The direct method requires careful counting, and is not recommended for general use:
𝐵𝑜𝑢𝑛𝑑𝑎𝑟𝑖𝑒𝑠 = ⏟ 5 + ⏟ 5 + ⏟ 4 + ⏟ 4 = 18
𝑇𝑜𝑝 𝐵𝑜𝑡𝑡𝑜𝑚 𝑅𝑖𝑔ℎ𝑡 𝐿𝑒𝑓𝑡

4.8 Sets, Relations and Functions

A. Subsets

4.192: Number of Subsets

The number of distinct subsets of a set with 𝑛 elements is
2𝑛

Making a subset is equivalent to choosing each element or not to put it in the subset.
For each element we have the same number of choices. Hence, this is multiplication principle with repetition.

⏟
2 × ⏟
2 ×…× ⏟
2 = 2𝑛
𝑭𝒊𝒓𝒔𝒕 𝑺𝒆𝒄𝒐𝒏𝒅 𝒏𝒕𝒉
𝑬𝒍𝒆𝒎𝒆𝒏𝒕 𝑬𝒍𝒆𝒎𝒆𝒏𝒕 𝑬𝒍𝒆𝒎𝒆𝒏𝒕

Example 4.193: Subsets

A set X has 2023 elements. What is the number of
A. distinct proper subsets of X?
B. of distinct non-null subsets of X?
C. of distinct, non-null, proper subsets of X?

Part A
We subtract the choice where we take all the elements.
22023 − 1
Part B
We subtract the choice where we take none of the elements.
22023 − 1
Part C
We subtract the choice where we take all of the elements, and also the choice where we take none of the
elements.
22023 − 1 − 1 = 22023 − 2

Example 4.194

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Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the
guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of
guides and tourists are possible? (AMC 10A 2004/12)

Let the tourists be

𝑇 = {𝑡1 , 𝑡2 , 𝑡3 , 𝑡4 , 𝑡5 , 𝑡6 }
Make a subset of T (𝑠𝑎𝑦 𝐺1 ) and assign all the tourists in the subset to the first guide, and send the remaining to
the second set.
⏟
2 × ⏟
2 × …× ⏟
2 = 26 = 64
𝐹𝑖𝑟𝑠𝑡 𝑆𝑒𝑐𝑜𝑛𝑑 𝑆𝑖𝑥𝑡ℎ
𝑇𝑜𝑢𝑟𝑖𝑠𝑡: 𝑇𝑜𝑢𝑟𝑖𝑠𝑡: 𝑇𝑜𝑢𝑟𝑖𝑠𝑡:
𝑇𝑤𝑜 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑇𝑤𝑜 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝑇𝑤𝑜 𝐶ℎ𝑜𝑖𝑐𝑒𝑠

However, we do not want the empty and the entire set, so subtract these two choices:
64 − 2 = 62
B. Cartesian Product

4.195: Cartesian Product

The Cartesian product of two sets 𝐴 and 𝐵 consists of the set of all 𝑜𝑟𝑑𝑒𝑟𝑒𝑑 𝑝𝑎𝑖𝑟s such that
➢ The first element is from Set 𝐴
➢ The second element is from Set 𝐵
Note that order matters: the first element cannot be from Set 𝐵.

Example 4.196
Consider the sets:
𝐴 = {𝑎, 𝑏}, 𝐵 = {𝑋, 𝑌, 𝑍}
A. Calculate the Cartesian product of 𝐴 and B.
B. Count the number of pairs listed in Part A.

We can form pairs by taking elements from the first set, and the second set like this:
{(𝑎, 𝑋), (𝑎, 𝑌), (𝑎, 𝑍), (𝑏, 𝑋), (𝑏, 𝑌), (𝑏, 𝑍)}
This is six ordered pairs.

Note that taking the first element from B, and the second element from A would have given us six more pairs,
like this:
{(𝑋, 𝑎), (𝑌, 𝑎), (𝑍, 𝑎), (𝑋, 𝑏), (𝑌, 𝑏), (𝑍, 𝑏)}
But this is against the condition given in the question that the first element must be from Set 𝐴.

Example 4.197
Consider the set of letters and the set of digits:
𝐿 = {𝐴, 𝐵, 𝐶, … , 𝑍}, 𝐷 = {0,1,2, … ,9}
Find the number of ordered pairs in the Cartesian Product of the above two sets.

26
⏟ × 10
⏟ = 260
𝑵𝒐. 𝒐𝒇 𝑵𝒐. 𝒐𝒇
𝑳𝒆𝒕𝒕𝒆𝒓𝒔 𝑫𝒊𝒈𝒊𝒕𝒔

(𝐴, 0), (𝐴, 1), … , (𝐴, 9) ⇒ 10 𝑃𝑎𝑖𝑟𝑠

(𝐵, 0), (𝐵, 1), … , (𝐵, 9) ⇒ 10 𝑃𝑎𝑖𝑟𝑠
.

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.
.
(𝑍, 0), (𝑍, 1), … , (𝑍, 9) ⇒ 10 𝑃𝑎𝑖𝑟𝑠

𝑇𝑜𝑡𝑎𝑙 = ⏟
10 + 10 + ⋯ 10 = 10 × 26 = 260
𝟐𝟔 𝑻𝒊𝒎𝒆𝒔

4.198: Cartesian Product: Number of Pairs

The number of pairs in the Cartesian Product of Set 𝐴 (with 𝑚 elements) and Set 𝐵 (with 𝑛 elements) is:
𝑚×𝑛

C. Relations

4.199: Relation
A relation between two sets 𝐴 and 𝐵 is a subset of the Cartesian product of the two sets.

Example 4.200
A set 𝑋 has 3 elements. A set 𝑌 has four elements. Find the number of relations between 𝑋 and 𝑌.

The Cartesian product has

3 × 4 = 12 𝑝𝑎𝑖𝑟𝑠

The number of relations is:

212 = 4096
D. Functions

4.201: Functions
A function is a relation from a set 𝑋 (called the domain) to a set 𝑌 (called the codomain) such that every
element in X is paired with exactly one element in Y.

It is written
𝑓: 𝑋 → 𝑌
Where
𝑋 𝑖𝑠 𝑑𝑜𝑚𝑎𝑖𝑛
𝑌 𝑖𝑠 𝑐𝑜𝑑𝑜𝑚𝑎𝑖𝑛

Example 4.202
A function 𝑓: ℕ → ℕ is defined as 𝑦 = 𝑓(𝑥) = √𝑥. For 0 ≤ 𝑥 ≤ 100, 𝑋 is the domain for 𝑓, and 𝑌 is the image of
𝑓. Find the number of elements in the Cartesian product of 𝑋 and 𝑌.

𝑋 = {12 , 22 , … , 102 }
𝑌 = {1,2, … ,10}

10 × 10 = 100

4.203: Injective Functions

An injective function is a function in which every element of the function’s codomain is the image of 𝑎𝑡 𝑚𝑜𝑠𝑡
one element of its domain.

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This means if the domain is X and the codomain is Y, then there is at most one value of 𝑦 ∈ 𝑌 such that
𝑦 = 𝑓(𝑥)

In other words, every input has a unique output.

Example 4.204
Find the number of injective functions such that:
𝑓: 𝑋 → 𝑌, 𝑛(𝑋) = 6, 𝑛(𝑌) = 4

For a function to be injective

𝑛(𝑋) ≤ 𝑛(𝑌) ⇒ 𝑁𝑜𝑡 𝑡𝑟𝑢𝑒
𝑁𝑜 𝑠𝑢𝑐ℎ 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠

Example 4.205
Find the number of injective functions such that
𝑓: 𝑋 → 𝑌, 𝑛(𝑋) = 4, 𝑛(𝑌) = 6

𝑋 ∈ {𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 }

6!
⏟
6 × ⏟
5 × ⏟
4 × ⏟
3 = = 6𝑃2
𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
2!
𝑓𝑜𝑟 𝑥1 𝑓𝑜𝑟 𝑥2 𝑓𝑜𝑟 𝑥3 𝑓𝑜𝑟 𝑥4

4.206: Surjective Functions

A surjective function is a function where every element of the function's codomain is the image of at least one
element of its domain.

Some different ways of looking at the same thing:

➢ If 𝑌 is the codomain, then
𝑦 ∈ 𝑌 ⇒ 𝑦 = 𝑓(𝑥) 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑥
➢ For a surjective function, the codomain is equal to the range.
➢ There is no element of the codomain which is not an output for some element in the domain.

Example 4.207
Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the
guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of
guides and tourists are possible? (AMC 10A 2004/12)

Let the tourists and guides be

𝑇 = {𝑡1 , 𝑡2 , 𝑡3 , 𝑡4 , 𝑡5 , 𝑡6 }, 𝐺 = {𝑔1 , 𝑔2 }

We want to assign each tourist to a guide. In other words, we want:

𝑔 = 𝑓(𝑡), 𝑓 𝑖𝑠 𝑠𝑢𝑟𝑗𝑒𝑐𝑡𝑖𝑣𝑒

⏟
2×⏟ 2 = 26 = 64
2 × …× ⏟
𝑡1 𝑡2 𝑡6

However, we do not want the cases where we assign a single output to every input since then the other

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element/guide in the codomain does not have a corresponding element/tourist in the domain.
𝑓(𝑡) = 𝑔1 , 𝑓(𝑡) = 𝑔2

Example 4.208
Find the number of surjective functions such that
𝑓: 𝑇 → 𝐺, 𝑛(𝑇) = 6, 𝑛(𝐺) = 2

As in the previous example:

26 − 2 = 62

Example 4.209
Find the number of surjective functions such that
𝑓: 𝑇 → 𝐺, 𝑛(𝑇) = 6, 𝑛(𝐺) = 3

If there are no restrictions on the function, the number of functions will be:
36 = 729

We want that every element in 𝐺 should be the output for at least one input. Hence, we need to subtract the
number of ways in which this is not fulfilled.

Case I: The number of functions where domain has 6 elements and image has exactly one element. In other
words, all inputs have a single output:
𝑓(𝑡) = 𝑔1 , 𝑓(𝑡) = 𝑔2 , 𝑓(𝑡) = 𝑔3 ⇒ 3 𝑤𝑎𝑦𝑠

Case II: If the set of all outputs has exactly two elements, the number of such functions is:
3
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑐ℎ𝑜𝑜𝑠𝑒 2 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑢𝑡 𝑜𝑓 3 = ( )
2

The number of functions where domain has 6 elements and image has exactly two elements:
26 − 2

3
( ) (26 − 2) = 3(62) = 186
2
The final answer is:
729 − 3 − 186 = 540

Example 4.210
Find the number of surjective functions such that
𝑓: 𝑇 → 𝐺, 𝑛(𝑇) = 6, 𝑛(𝐺) = 4

If there are no restrictions on the function, the number of functions will be:
46 = 212 = 4096

We want that every element in 𝐺 should be the output for at least one input. Hence, we need to subtract the
number of ways in which this is not fulfilled.

Case I: The number of functions where domain has 6 elements and image has exactly one element. In other
words, all inputs have a single output:
4 𝑤𝑎𝑦𝑠

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Case II: If the set of all outputs has exactly two elements, the number of such functions is:
4
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑐ℎ𝑜𝑜𝑠𝑒 2 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑢𝑡 𝑜𝑓 3 = ( )
2

The number of functions where domain has 6 elements and image has exactly two elements:
26 − 2

4
( ) (26 − 2) = 6(62) = 372
2

Case II: If the set of all outputs has exactly three elements, the number of such functions is:
4
( ) [540] = 4(540) = 2160
3

The final answer is:

4096 − 4 − 372 − 2160 = 1560

4.211: Bijective Functions

A function is bijective if it is both injective and surjective.

In other words, every input has exactly one output, and every output is associated with exactly one input.

Example 4.212
Find the number of bijective functions such that
𝑓: 𝑇 → 𝐺, 𝑛(𝑇) = 7, 𝑛(𝐺) = 5

𝑁𝑜 𝑠𝑢𝑐ℎ 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠

Example 4.213
Find the number of surjective functions such that
𝑓: 𝑇 → 𝐺, 𝑛(𝑇) = 6, 𝑛(𝐺) = 6

𝑇 = {𝑡1 , 𝑡2 , 𝑡3 , 𝑡4 , 𝑡5 , 𝑡6 }
⏟
6 × ⏟5 × …× ⏟ 1 ⇒ 6! = 720
𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝐶ℎ𝑜𝑖𝑐𝑒𝑠 𝐶ℎ𝑜𝑖𝑐𝑒𝑠
𝑓𝑜𝑟 𝑡1 𝑓𝑜𝑟 𝑡2 𝑓𝑜𝑟 𝑡6

E. Computer Science

Example 4.214: Values in a Bit

Signed Bit
Unsigned Bit

Example 4.215: IPv4

IPv4
IPV6

216 Examples

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