YEAR 10
PERMUTATION AND
COMBINATION
TERM 4
2023-2024
01
ARRANGEMENT
FACTORIAL NOTATION
✏The multiplication of all positive integers, say “n”, that will be
smaller than or equivalent to n. For any integer 𝑛 > 0,
𝒏! = 𝒏 𝒏 − 𝟏 𝒏 − 𝟐 … × 𝟑 × 𝟐 × 𝟏
Note: 0! = 1
Example
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
6!
=
4!
5! − 4! =
3!
=
0!
Example
Find the value of:
5! 10! 9!
a. d. +
3! 8! 7!
4! 20! 13!
b. − 3! e. −
2! 18! 11!
c. 7 × 4! + 21 × 3!
Example
Rewrite each of the following using
factorial notation.
a. 2×1
b. 5 × 4 × 3
10×9×8
c.
4×3×2×1
ARRANGEMENT
These books are arranged in the order GORB. (Green, Orange,
Red, Blue). The books could be arranged in the order OGBR.
Find the number of different ways that the 4 books can be arranged
in a line. You will need to be systematic.
How many ways are there of arranging five different books in a line?
The number of ways of arranging 𝑛 distinct items
in a line:
n!
Example
G R A D I E N T
a. Find the number of different arrangements of these nine
cards if there are no restrictions.
b. Find the number of arrangements that begin with GRAD.
c. Find the number of arrangements that begin with G and end
with S.
Solution
a. There are 9 different cards. So, the number of
arrangements = 9! = 362 880.
b. The first four letters are GRAD, so there are now only 5
letters left to be arranged.
Number of arrangements = 5! = 120
c. The first and last letters are fixed, so there are now 7
letters to arrange between the G and the S.
number of arrangements = 7! = 5040
Example
a. Find the number of different arrangements of these
seven objects if there are no restrictions.
b. Find the number of arrangements where the
squares and circles alternate.
c. Find the number of arrangements where all the
squares are together.
d. Find the number of arrangements where the
squares are together and the circles are together.
02
PERMUTATIONS
Permutations of n distinct objects
The number of permutation of 𝑛 distinct objects is denoted by
𝑛
𝑃𝑛 , and there are 𝑛! Permutation that can be made.
𝒏
𝑷𝒏 = 𝒏!
for any integer 𝑛 > 0
Note: In permutations, order matters
Example
Find the number of 3-digit numbers that can be made by
arranging the digits 5, 6, and 7.
There are 3 × 2 × 1 = 3! = 3𝑃3 = 6 ways
Permutations of n objects with
repetitions
The number of permutation of 𝑛 objects, of which 𝑝 are of one
type, 𝑞 are of another type, 𝑟 are of another type, and so on, is:
𝒏
𝑷𝒏 𝒏!
=
𝒑! × 𝒒! × 𝒓! × ⋯ 𝒑! × 𝒒! × 𝒓! × ⋯
✏where 𝑝+𝑞+𝑟+⋯=𝑛
Example
The capital of Burkina Faso is OUAGADOUGOU. Find the
number of distinct arrangements of all the letters in this
word.
Answer:
11!
= 277 200
3! × 3! × 2! × 2!
Note: 11 letters are to be arranged, with repeats of three
Os, three Us, two As, and two Gs.
Permutations of r objects from n
objects
𝒏 𝒏!
𝑷𝒓 =
𝒏−𝒓 !
Example
How many three-digit numbers can be made from the
seven digits 3, 4, 5, 6, 7, 8, and 9, if each is used at
most once?
Answer:
7 7! 7!
𝑃3 = = = 7 × 6 × 5 = 210
7 − 3 ! 4!
Example
In how many ways can five playing cards from a standard
deck of 52 cards be arranged in a row?
Example
In how many ways can 4 out of 18 girls sit on a four-seat
sofa when the oldest girl must be given one of the seats?
Answer:
4 × 17𝑃3 = 16 320 ways
Note: four ways for the oldest girl to occupy a seat, and 17
𝑃3 ways to
select and arrange three remaining 17 girls to sit with her.
Example
In how many ways can four boys and three girls stand in a row
when no two girls are allowed to stand next to each other?
Answer:
4
𝑃4 ways to arrange 4 boys in a row
𝑩𝟏 𝑩𝟐 𝑩𝟑 𝑩𝟒
5
Arrange the girls in 3 of these 5 spaces 𝑃3
4 5
𝑃4 × 𝑃3 = 1440 ways
Example
Find how many even numbers between 3000 and 4000
can be formed using the digits 1, 3, 5, 6, 7, and 9 of no
number can be repeated.
Solution
✏ Method 1
The first number must be a 3 and the last number must be a 6.
There are now two spaces to fill using two of the remaining four digits 1, 5, 7, 9.
4! 4!
Number of ways of filling the remaining two spaces = 4𝑃2 = = = 12.
4−2 ! 2!
✏ Method 2
Consider the number of choices for filling each of the four spaces.
Number of ways of filling the four spaces = 1 × 4 × 3 × 1
✏ Method 3
In this example it is not impractical to list all the possible permutations.
These are:
3156 3516 3176 3716 3196 3916
3576 3756 3196 3916 3796 3976
There are 12 different numbers that satisfy the conditions.
03
COMBINATIONS
Combination
A combination is simply a selection, where the order of selection is
not important. Choosing strawberries and ice cream from the menu
is the same combination as choosing ice cream and strawberries.
When we select 𝑟 objects in no particular order from 𝑛 objects, we
call this is a combination.
𝒏 𝒏 𝒏!
𝑪𝒓 = =
𝒓 𝒓! 𝒏 − 𝒓 !
Example
In how many ways can three fish be selected from a bowl
containing seven fish and two potatoes?
Answer:
7 7!
𝐶3 = = 35
3! × 4!
Example
In how many ways can five books and three magazines be
selected from eight books and six magazines?
Answer:
8
𝐶5 × 6𝐶3 = 56 × 20 = 1120
Example
A team of five is to be chosen from six women and five
men. Find the number of possible teams in which there
will be more women than men.
Answer:
From 6 women From 5 men Number of teams
6
3 2 𝐶3 × 5𝐶2 = 20 × 10 = 200
6
4 1 𝐶4 × 5𝐶1 = 15 × 5 = 75
6
5 0 𝐶5 × 5𝐶0 = 6 × 1 = 6
200 + 75 + 6 = 281 teams with more women than men.
Example
How many distinct three-digit numbers can be made from
five cards, each with one of the digits 5, 5, 7, 8, and 9
written on it?
Solution
The 5 is repeated digit, so we must investigate three situations
separately.
No 5s selected:
3
𝑃3 = 6 three-digit numbers
One 5 selected:
3
𝐶2 × 3! = 18 three-digit numbers
Two 5s selected:
3!
3
𝐶1 × = 9 three-digit numbers
2!
6 + 18 + 9 = 33 three-digit numbers can be made.
Example
Sofia has to play 5 pieces of music for her music
examination.
She has 13 pieces of music to choose from.
There are 7 pieces written by Chopin, 4 written by Mozart
and 2 written by Bach.
Find the number of ways the 5 pieces can be chosen if:
a. There are no restriction
b. There must be 2 pieces by Chopin, 2 pieces by
Mozart and 1 piece by Bach
c. There must be at least one piece by each composer
Solution
a. Number of ways of choosing 5 from 13 = 13𝐶5 = 1287
b. Number of ways of choosing 2 from 7 pieces by Chopin
= 𝟕𝑪𝟐
Number of ways of choosing 2 from 4 pieces by Mozart = 𝟒𝑪𝟐
Number of ways of choosing 1 from 2 pieces by Bach = 𝟐𝑪𝟏
So, number of possible selections = 𝟕𝑪𝟐 × 𝟒𝑪𝟐 × 𝟐𝑪𝟏 = 21
× 6 × 2 = 252
Solution
c. If there is at least one piece by each composer there could be:
Chopin Mozart Bach Number of ways
7
3 1 1 𝑪3 × 4𝑪1 × 2𝑪1 = 35 × 4 × 2 = 280
7
1 3 1 𝑪1 × 4𝑪3 × 2𝑪1 = 7 × 4 × 2 = 56
7
2 2 1 𝑪2 × 4𝑪2 × 2𝑪1 = 21 × 6 × 2 = 252
7
2 1 2 𝑪2 × 4𝑪1 × 2𝑪2 = 21 × 4 × 1 = 84
7
1 2 2 𝑪1 × 4𝑪2 × 2𝑪2 = 7 × 6 × 1 = 42
So, total number of ways = 280 + 56 + 252 + 84 + 42 = 714
NOTE
Permutation Combination
• A permutation is a way of • A combination is a way of
selecting and arranging objects in selecting objects in no particular
a particular order. order.
• Order is important. • Order is not important.
• A key word that points to a • Key words that point to
permutation is arranged. combination are chosen and
• Arranging people, digits, selected.
numbers, alphabets, letters, • Selection of menu, food, clothes,
colours, etc. subjects, teams, etc.
