Permutation is an arrangement of objects in a definite order. When we look at the schedules of trains, buses and the flights we really wonder how they are scheduled according to the public’s convenience. Of course, the permutation is very much helpful to prepare the schedules on departure and arrival of these. Also when we come across license plates of vehicles which consists of few alphabets and digits. We can easily prepare these codes using permutations.

Also, read: Permutation And Combination

Table of contents:

Permutation in Maths

Definition of Permutation

Basically Permutation is an arrangement of objects in a particular way or order. While dealing with permutation one should concern about the selection as well as arrangement. In Short, Ordering is very much essential in permutations.

Representation of Permutation

We can represent permutation in many ways, such as:

  • \(\large \mathbf{P(n,k)}\)
  • \(\large \mathbf{P^{n}_{k}}\)
  • \(\large \mathbf{_{n}P_{k}}\)
  • \(\large \mathbf{^{n}P_{k}}\)
  • \(\large \mathbf{P _{n}\, _{,k}}\)

Types of Permutation

Permutation can be classified in three different categories:

  • Permutation of n different objects ( when repetition is not allowed)
  • Repetition, where repetition is allowed
  • Permutation when the objects are not distinct (Permutation of multisets)

Let us understand all the cases of permutation in details.

Permutation of n different objects

If n is a positive integer and r is a whole number, such that r < n, then P(n, r) represents the number of all possible arrangements or permutations of n distinct objects taken r at a time. It can also be represented as


P(n, r) = n(n-1)(n-2)(n-3)……..upto r factors

\(\Rightarrow\) P(n, r) = n(n-1)(n-2)(n-3)……..(n – r +1)

\(\large \Rightarrow P(n,r) = \frac{n!}{(n-r)!}\)

Example: How many 3 letter words with or without meaning can be formed out of the letters of the word SWING when repetition of letters is not allowed?

Solution: Here n = 5, as the word SWING has 5 letters. Since we have to frame 3 letter words with or without meaning and without repetition, therefore total permutations possible are:

\(\large \Rightarrow P(n,r) = \frac{5!}{(5-3)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 60\)

Permutation when repetition is not allowed

When the number of object is “n,” and we have “r” to be the selection of object, then

Choosing an object can be in n different ways (each time).

Thus the permutation of objects when repetition is allowed will be equal to,

n × n × n × ……(r times) = nr

Example: How many 3 letter words with or without meaning can be formed out of the letters of the word SMOKE when repetition of words is allowed?


The number of objects, in this case, is 5, as the word SMOKE has 5 alphabets.

and r = 3, as 3 letter word has to be chosen.

Thus the permutation will be

Permutation (when repetition is allowed) = \(\large 5^{3}\)

\(\large = 125\)

Permutation of multi-sets

Permutation of n different objects when \(p_{1}\) objects among ‘n’ objects are similar, \(p_{2}\) objects of the second kind are similar, \(p_{3}\) objects of the third kind are similar ……… and so on, \(p_{k}\) objects of the kth kind are similar and the remaining of all are of a different kind,

Thus it forms a multiset, where the permutation is given as:

\(\large \mathbf{\large \frac{n!}{p_{1}!\; p_{2}!\; p_{3}…..p_{n}!}}\)

Fundamental Counting Principle

According to this principle, “If one operation can be performed in ‘m’ ways and there are n ways of performing a second operation, then the number of ways of performing the two operations together is m x n “.

This principle can be extended to the case in which the different operation be performed in m, n, p, . . . . . . ways.

In this case the number of ways of performing all the operations one after the other is m x n x p x . . . . . . . . and so on

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Examples on Permutation

Example 1: In how many ways 6 children can be arranged in a line, such that

(i) Two particular children of them are always together

(ii) Two particular children of them are never together


(i) The given condition states that, 2 students needs to be together, hence we can consider them 1.

Thus, the remaining 7 gives the arrangement in 5! ways, i.e. 120.

Also the two children in a line can be arranged in 2! Ways.

Hence, the total number of arrangements will be,

5! × 2! = 120 × 2 = 240 ways

(ii) The total number of arrangements of 6 children will be 6!, i.e. 720 ways.

Out of the total arrangement, we know that, two particular children when together can be arranged in 240 ways.

Therefore, total arrangement of children in which two particular children are never together will be 720 – 240 ways, i.e. 480 ways.

Example 2:Consider a set having 5 elements a,b,c,d,e. In how many ways 3 elements can be selected (without repetition) out of the total number of elements.

Solution: Given X = {a,b,c,d,e}

3 are to be selected.

Therefore, \(P_{3}^{5} = \frac{5!}{(5-3)!}\)

= 60

Example 3: It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

Solution: We are given that there are 5 men and 4 women.

i.e. there are 9 positions.

The even positions are , 2nd, 4th, 6th and the 8th places and

These four places can be occupied by 4 women in P(4, 4) ways = 4! 

= 4 . 3. 2. 1 

= 24 ways

The remaining 5 positions can be occupied by 5 men in P(5, 5) = 5! 


= 120 ways

Therefore, by the Fundamental Counting Principle, 

Total number of ways of seating arrangements = 24 x 120 

= 2880

Practice Problems

Practice below listed problems:

  1. How many numbers lying between 100 and 1000 can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed.
  2. Seven athletes are participating in a race. In how many ways can the first three prizes be won.

To solve more problems or to take a test, download BYJU’S – The Learning App from Google Play Store.

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