Permutation Formula

In mathematics, permutation relates to the method of organizing all the members of a group into some series or design. In further terms, if the group is already completed, then the redirecting of its components is called the method of permuting. Permutations take place, in better or slightly effective methods, in almost every district of mathematics. They usually happen when different direction on detailed restricted sites is monitored.

Permutation formula

It is the separate arrangements of a supplied numeral of associates taken one by one, or some, or all at a time. For instance, if we have two elements A and B, then there are two possible interpretations, AB and BA.

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

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.

Explanation of Permutation formula

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)!

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 phrases, with or without meaning, that can be comprised with the letters of the word ‘TABLE’.

Solution:

‘TABLE’ contains 5 letters.

Thus, the numeral of phrases that can be formed with these 5 letters = 5! = 5 × 4 × 3 × 2 × 1 = 120.

Question 7: 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.