PERMUTATION

At the end of the lesson, the learners

should be able to
• Illustrate the permutation of objects
• Derive the formula for finding the
number of permutations of 𝑛
objects taken 𝑟 at a time
• Solve problems involving
permutations.
 PERMUTATION

It is the arrangement of objects in a set. The

order of objects is important in arrangement.

Example: abc and acb

 Factorial Notation

FACTORIAL refers to the product of all positive integers less

than or equal to a particular positive integer. It is usually
denoted by that integer and an exclamation point.

𝑛! = 𝑛 × 𝑛 − 1 × … × 2 × 1

NOTE: 0! = 1
Factorial Notation

Example:
Evaluate 3!.
Factorial Notation

Example:
To evaluate 3!, multiply the natural numbers in decreasing
order starting from 3.

3! = 3 × 2 × 1
=6
 Ways on how to illustrate permutation of objects.

1. Fundamental Principle of Counting

2. TREE Diagram

3. Listing Method
 Fundamental Counting Principle

This is a mathematical rule that allows you to

find the number of ways that a combination of
events can occur. If there p ways to do one
thing and q ways to do another thing, then
there are 𝑝 × 𝑞 ways to do both.
Fundamental Counting Principle

Example:
In how many ways can you arrange the letter P, Q, R, and S?
 RULES IN PERMUTATION

• LINEAR PERMUTATION
• PERMUTATION OF 𝑛 OBJECTS TAKEN AT 𝑟
• PERMUTATION WITH REPETITION
• PERMUTATION OF THINGS THAT ARE
ALIKE
• CIRCULAR PERMUTATION
 LINEAR PERMUTATION

It is an ordered arrangement of objects in a line. The

number of linear permutations of 𝒏 objects is computed
using the rule 𝒏 𝑷𝒏 = 𝒏!
 LINEAR PERMUTATION

Example:
Rodo is going to arrange his textbooks on a shelf: Algebra,
Calculus, Geometry, Statistics and Trigonometry. How
many different possible arrangement can he make?
LINEAR PERMUTATION
Example:
Rodo is going to arrange his textbooks on a shelf: Algebra, Calculus, Geometry, Statistics and Trigonometry.
How many different possible arrangement can he make?

Solution:
The total number of books needs to arrange is 𝟓. So, by using
the linear permutation rule, the total number of possible
arrangement is:

𝟓! = 𝟓 × 𝟒 × 𝟑 × 𝟐 × 𝟏
= 𝟏𝟐𝟎
 LINEAR PERMUTATION

More Examples:
1. In how many ways can 7 basketball players be seated on
a bench?
2. In how many ways can 9 persons be arranged in a row of
9 chairs.
 PERMUTATION OF 𝑛 OBJECTS TAKEN AT 𝑟

The number of permutations of 𝒏 objects taken

𝒏!
by 𝒓 at time is 𝒏 𝑷𝒓 = where 𝒏 is the total
𝒏−𝒓 !
number of objects and 𝒓 is the number of objects
taken at a time.
 PERMUTATION OF 𝑛 OBJECTS TAKEN AT 𝑟

Example:
Lynzie, Elisha, Angel, Shaina, and Kyla wants to
take a photo in which three of the five friends
are lined up in a row. How many different photos
are possible?
PERMUTATION OF 𝑛 OBJECTS TAKEN AT 𝑟
Example:
Lynzie, Elisha, Angel, Shaina, and Kyla wants to take a photo in which three of the five friends are lined up
in a row. How many different photos are possible?

Solution: 𝒏! 𝟓×𝟒×𝟑×𝟐×𝟏
𝒏 =? 𝒏 𝑷𝒓 = =
𝒏−𝒓 ! 𝟐×𝟏
𝒓 =?
𝟓!
𝟓 𝑷𝟑 =
𝒏=𝟓 𝟓−𝟑 ! 𝟓 𝑷𝟑 = 𝟔𝟎
𝒓=𝟑 𝟓!
=
𝟐!
 PERMUTATION OF 𝑛 OBJECTS TAKEN AT 𝑟

More Example:
Calculate the number of permutations if
there are 6 distinct objects taken 3 at a
time.
PERMUTATION OF 𝑛 OBJECTS TAKEN AT 𝑟
Example:
Calculate the number of permutations if there are 6 distinct objects taken 3 at a time.

Solution: 𝒏! 𝟔×𝟓×𝟒×𝟑×𝟐×𝟏
𝒏 =? 𝒏 𝑷𝒓 = =
𝒏−𝒓 ! 𝟑×𝟐×𝟏
𝒓 =?
𝟔!
𝟔 𝑷𝟑 =
𝒏=𝟔 𝟔−𝟑 ! 𝟔 𝑷𝟑 = 𝟏𝟐𝟎
𝒓=𝟑 𝟔!
=
𝟑!
 PERMUTATION OF 𝑛 OBJECTS TAKEN AT 𝑟

More Example:
1. Calculate the number of permutations of 𝟕 different plants
taken 3 at a time.
2. There are 10 passengers in a bus terminal going to Baguio
city. How many ways can 3 passengers be arranged out of
10 on a queue
 PERMUTATION

More Example:

A family of 7 wants to watch a movie in a cinema together.

1. In how many different ways can they be arranged in a row of
7 seats?
2. In how many ways can they be arranged in a row of seven
seats if there are 2 kids who need to seated together?
PERMUTATION
Example:
A family of 7 wants to watch a movie in a cinema together.
1. In how many different ways can they be arranged in a row of 7 seats?

Solution: 𝒏! 𝟕×𝟔×𝟓×𝟒×𝟑×𝟐×𝟏
𝒏 =? 𝒏 𝑷𝒓 = =
𝒏−𝒓 ! 𝟏
𝒓 =?
𝟕!
𝟕 𝑷𝟕 =
𝒏=𝟕 𝟕−𝟕 ! 𝟕 𝑷𝟕 = 𝟓𝟎𝟒𝟎
𝒓=𝟕 𝟕!
=
𝟎!
PERMUTATION
Example:
A family of 7 wants to watch a movie in a cinema together.
1. In how many ways can they be arranged in a row of seven seats if there are 2 kids who
need to seated together?

Solution: 𝒏! 𝟕×𝟔×𝟓×𝟒×𝟑×𝟐×𝟏
𝒏 =? 𝒏 𝑷𝒓 = =
𝒏−𝒓 ! 𝟓×𝟒×𝟑×𝟐×𝟏
𝒓 =?
𝟕!
𝟕 𝑷𝟐 =
𝒏=𝟕 𝟕−𝟐 ! 𝟕 𝑷𝟐 = 𝟒𝟐
𝒓=𝟐 𝟕!
=
𝟓!
 PERMUTATION

ASSIGNMENT:
1. Find the number of ways by which 5 members from a family
of 7 can line up in a photo booth.
2. Jane has to finish 6 assignments this weekend.
a. In how many different orders can she complete the
assignments?
b. In how many ways can she choose 2 of the assignments at a
time?
PERMUTATION
Assignment:
1. Find the number of ways by which 5 members from a family of 7 can line up in a photo
booth.

Solution: 𝒏! 𝟕×𝟔×𝟓×𝟒×𝟑×𝟐×𝟏
𝒏 =? 𝒏 𝑷𝒓 = =
𝒏−𝒓 ! 𝟐×𝟏
𝒓 =?
𝟕!
𝟕 𝑷𝟓 =
𝒏=𝟕 𝟕−𝟓 ! 𝟕 𝑷𝟓 = 𝟐, 𝟓𝟐𝟎
𝒓=𝟓 𝟕!
=
𝟐!
PERMUTATION
Assignment:
Jane has to finish 6 assignments this weekend.
a. In how many different orders can she complete the assignments?

Solution: 𝒏! 𝟔×𝟓×𝟒×𝟑×𝟐×𝟏
𝒏 =? 𝒏 𝑷𝒓 = =
𝒏−𝒓 ! 𝟏
𝒓 =?
𝟔!
𝟔 𝑷𝟔 =
𝒏=𝟔 𝟔−𝟔 ! 𝟔 𝑷𝟔 = 𝟕𝟐𝟎
𝒓=𝟔 𝟔!
=
𝟏!
PERMUTATION
Assignment:
Jane has to finish 6 assignments this weekend.
a. In how many ways can she choose 2 of the assignments at a time?

Solution: 𝒏! 𝟔×𝟓×𝟒×𝟑×𝟐×𝟏
𝒏 =? 𝒏 𝑷𝒓 = =
𝒏−𝒓 ! 𝟒×𝟑×𝟐×𝟏
𝒓 =?
𝟔!
𝟔 𝑷𝟐 =
𝒏=𝟔 𝟔−𝟐 ! 𝟔 𝑷𝟐 = 𝟑𝟎
𝒓=𝟐 𝟔!
=
𝟒!
PERMUTATION WITH REPETITION

The number of permutations of 𝒏 different

objects taken 𝒓 at time is given as 𝑷 = 𝒏𝒓
Where:
𝑷 is the total number of permutations
𝒏 is the total number of objects in a set
𝒓 is the total number of choosing objects in a set
PERMUTATION WITH REPETITION

Example:
1. How many 𝟓 − letter ordered codes can be
formed with the letters of the set {𝑨, 𝑩, 𝑪, 𝑫} if we
allow to use repeated use of the same letter?
PERMUTATION WITH REPETITION
Example:
1. How many 𝟓 − letter ordered codes can be formed with the letters of the set
{𝑨, 𝑩, 𝑪, 𝑫} if we allow to use repeated use of the same letter?

Formula: 𝑷 = 𝒏𝒓
Solution: 𝑷 = 𝒏𝒓
𝒏=?
𝒓=? 𝑷 = (𝟒)𝟓
𝒏=𝟒 𝑷=𝟒×𝟒×𝟒×𝟒×𝟒
𝒓=𝟓
𝑷 = 𝟏𝟎𝟐𝟒
PERMUTATION WITH REPETITION

Example:
2. A standard deck of cards has 52 different
cards. How many 𝟐 − cards ordered arrangements
can be made by selecting 2 cards with
replacement?
PERMUTATION WITH REPETITION
Example:
A standard deck of cards has 52 different cards. How many 𝟐 − cards ordered
arrangements can be made by selecting 2 cards with replacement?

Formula: 𝑷 = 𝒏𝒓
Solution: 𝑷 = 𝒏𝒓
𝒏=?
𝒓=? 𝑷 = (𝟓𝟐)𝟐
𝒏 = 𝟓𝟐 𝑷 = 𝟓𝟐 × 𝟓𝟐
𝒓=𝟐 𝑷 = 𝟐𝟕𝟎𝟒
𝟐 − card ordered arrangements
 PERMUTATION WITH REPETITION

Example:
3. The standard configuration for a student identification
(ID) number in a certain public schools is 2 digits followed by 2
letters.
a. How many different student ID numbers are possible if the
digits and letters can be repeated?
b. How many different student ID numbers are possible if the
digits and letters cannot be repeated?
PERMUTATION WITH REPETITION
Example:
The standard configuration for a student identification (ID) number in a certain public schools is 2
digits followed by 2 letters. How many different student ID numbers are possible if the digits and letters can be
repeated?

Formula: 𝑷 = 𝒏𝒓
Solution: 𝑷 = 𝒏𝒓 ∙ 𝒍𝒓
𝒏=?
𝒓=? 𝑷 = (𝟏𝟎)𝟐 ∙ (𝟐𝟔)𝟐
𝒏 = 𝟏𝟎 and 𝟐𝟔 𝑷 = 𝟏𝟎𝟎 × 𝟔𝟕𝟔
𝒓=𝟐 𝑷 = 𝟔𝟕, 𝟔𝟎𝟎
𝟐 − card ordered arrangements
PERMUTATION WITH REPETITION
Example:
The standard configuration for a student identification (ID) number in a certain public schools is 2
digits followed by 2 letters. How many different student ID numbers are possible if the digits and letters cannot
be repeated?

Formula: 𝑷 = 𝒏𝒓
Solution: 𝑷 = 𝒏𝒓 ∙ 𝒍𝒓
𝒏=?
𝒓=? 𝑷 = (𝟏𝟎)𝟐 ∙ (𝟐𝟔)𝟐
𝒏 = 𝟏𝟎 and 𝟐𝟔 𝑷 = 𝟏𝟎𝟎 × 𝟔𝟕𝟔
𝒓=𝟐 𝑷 = 𝟔𝟕, 𝟔𝟎𝟎
𝟐 − card ordered arrangements
PERMUTATION OF THE THINGS THAT ARE ALIKE

The number of distinct permutations of 𝒏 objects

where 𝒏𝟏 are alike, 𝒏𝟐 are alike, 𝒏𝟑 are alike…, 𝒏𝒌
𝒏!
is defined as .
𝒏𝟏 !𝒏𝟐 !𝒏𝟑 !…𝒏𝒌
PERMUTATION OF THE THINGS THAT ARE ALIKE

Example:
1. In how many ways can the letters of the word
CALCULUS be arranged?
PERMUTATION OF THE THINGS THAT ARE ALIKE
Example:
1. In how many ways can the letters of the word CALCULUS be arranged?

Solution: 𝑻𝒉𝒆 𝒕𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒍𝒆𝒕𝒕𝒆𝒓 𝒊𝒏 𝒘𝒐𝒓𝒅 𝑪𝑨𝑳𝑪𝑼𝑳𝑼𝑺 𝒊𝒔 𝟖 that

is 𝐧 = 𝟖, 𝒘𝒉𝒆𝒓𝒆 𝒕𝒉𝒆𝒓𝒆 𝒂𝒓𝒆 𝟐𝑪𝒔, 𝟏 𝑨, 𝟐𝑳𝒔, 𝟐 𝑼𝒔 𝒂𝒏𝒅 𝟏 𝑺.
𝟖! 𝟖×𝟕×𝟔×𝟓×𝟒×𝟑×𝟐×𝟏
=
𝟐! 𝟏! 𝟐! 𝟐! 𝟏! (𝟐 × 𝟏)(𝟏)(𝟐 × 𝟏)(𝟐 × 𝟏)(𝟏)

𝟒𝟎, 𝟑𝟐𝟎 𝟐𝟎, 𝟏𝟔𝟎

= = = 𝟓𝟎𝟒𝟎
(𝟏)(𝟐)(𝟐)(𝟏) 𝟒