Permutation Formula

Last Updated : 17 Jan, 2025

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.

Permutation Formulas

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

An integer of permutations when ‘r’ elements are arranged out of a total of ‘n’ elements is given by ,

ⁿP_r = n! / (n - r)!.
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 on Permutation Formula

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

The permutation of a collection of things or components in order relies on three conditions:
When recurrence of essences is not allowed
When recurrence of essences is allowed
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.
There are 6 consonants and 1 vowels in it.
No. of ways 1 vowels can occur in 7 different places = ⁷P₁ = 7 ways.
After 1 vowels take 1 place, no. of ways 6 consonants can take 6 places = ⁶P₆ = 6! = 720 ways.
Therefore, total number of permutations possible = 720 × 720 = 518,400 ways.

Also Check:

Permutation Formula

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