Permutation Formula

Permutation is a fundamental concept in mathematics that deals with arranging a set of objects in a specific order. It focuses on the different ways items can be ordered when the arrangement matters. Permutations are widely used in various fields, including probability, computer science, and data analysis, to solve problems like task scheduling, sequence generation, and optimization.

The permutation formula calculates the number of ways to arrange r objects from a set of n distinct objects, where order matters.

The number of permutations when ‘r’ elements are arranged out of a total of ‘n’ elements is given by:

For example : Let n = 2 (A and B) and r = 1 (All permutations of size 1). The answer is 2!/(2 - 1)! = 2. The two permutations are AB, and BA.

Permutation Formula Explanation

A permutation is a kind of arrangement that shows how to permute. If there are three separate integers 1, 2, and 3, and if somebody is interested to permute the integers taking 2 at a point, it offers (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), and (3, 2). That is it can be performed in 6 ways.

Here, (1, 2) and (2, 1) are separate. Again, if these 3 integers shall be set enduring all at a time, then the arrangements will be (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) and (3, 2, 1) i.e. in 6 ways.

In known, n separate items can be selected accepting r (r < n) at a time in n(n - 1)(n - 2) ... (n - r + 1) ways. In particular, the first item can be any of the n items. Now, after selecting the first item, the second item will be any of the remaining n - 1 thing. Similarly, the third item can be any of the remaining n - 2 things. Alike, the r^th item can be any of the remaining n - (r - 1) things.

Therefore, the total numeral of permutations of n separate items taking r at a time is n(n - 1)(n - 2) ... [n - (r - 1)] which is noted as ⁿP_r. Or, in other words,

ⁿP_r = n!/(n - r)!

Other Permutation Formulas

• ⁿp_n = n!

• ⁿp₀ = 1

• ⁿp₁ = n

• ⁿp_{n -1} = n

• ⁿp_r = n × ^{n - 1}p_{r -1} = n × (n - 1) × ^{n - 2}p_{r -2}

• ^{n - 1}p_r + r × ^{n - 1}p_{r -1} = ⁿP_r

• \frac{^np_r}{^np_{r-1}} = n - r + 1

• ⁿP_r = n! ( When repetition is allowed )

• Permutation of n items when q items are alike : n!/q!

Sample Problems

Question 1: What are the types of permutations?
Solution:

The permutation of a collection of things or components in order relies on three conditions:

1. When recurrence of essences is not allowed
2. When recurrence of essences is allowed
3. When the components of a group are not different

Question 2: Calculate the number of permutations of n = 5 and r = 2.
Solution:

Given,
n = 5
r = 2

Using the formula given above:

Permutation: ⁿP_r = (n!) / (n - r)!

= (5!) / (5 - 2)!
= 5! / 3! = (5 × 4 × 3! )/ 3!
= 20

Question 3: How many 3 letter phrases with or without purpose can be created out of the letters of the word POEM when repetition of letters is not permitted?

Solution:

Here n = 4, as the word POEM has 4 letters. Since we have to create 3 letter words with or without meaning and without repetition, therefore total permutations possible are:

⇒ P(n, r) = 4!/(4 − 3)!
= 4 × 3 × 2 × 1/1
= 24

Question 4: How many 4 letter phrases with or without purpose can be created out of the letters of the word KANHA when repetition of words is permitted?

Solution:

The number of letters, in this case, is 5, as the word KANHA has 5 alphabets.

And r = 4, as a 4-letter term has to be selected.

Thus, the permutation will be:
Permutation (when repetition is permitted) = 5⁴= 625

Question 5: It is required to place 4 men and 3 women in a row so that the women entertain the even positions. How many such configurations are feasible?
Solution: 

We are given that there are 4 men and 3 women.

i.e. there are 7 positions.

The even positions are: 2nd, 4th, and the 6th places
These three places can be occupied by 3 women in P(3, 3) ways = 3!  
= 3 × 2 × 1  
= 6 ways

The remaining 4 positions can be occupied by 4 men in P(4, 4) = 4!  

= 4 × 3 × 2 × 1  
= 24 ways

Therefore, by the Fundamental Counting Principle,  

Total number of ways of seating arrangements = 24 ×  6 
= 144

Question 6: Find the number of permutations of the letters of the phrase subject such that the vowels consistently appear in odd positions.
Solution:

The word “SUBJECT” has 7 letters.
Vowels = U, E = 2 vowels
Consonants = S, B, J, C, T = 5 consonants

Odd positions are 1, 3, 5, and 7.

Ways to place 2 vowels in 4 odd positions = ⁴P₂ = 4 × 3 = 12

Ways to arrange 5 consonants in remaining 5 places = 5! = 120

Therefore, total permutations = 12 × 120 = 1440

Hence, the correct answer is 1440.

Practice Problem

Question 1. How many ways can 3 books be arranged on a shelf if there are 5 books available to choose from?

Question 2. A teacher wants to select 4 students from a class of 10 to present a project, and the order in which they present matters. How many different ways can the teacher choose and arrange the students?

Question 3. How many 3-letter words can be formed from the letters of the word "MATH" when repetition of letters is allowed?

Question 4. How many distinct permutations can be made from the word "MISSISSIPPI"?

Answer:-

1. 20.
2. 5040.
3. 64.
4. 34650.