Permutation is an arrangement of objects in a definite order.


Representation of Permutation:

We can represent permutation in many ways,

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

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

\(\large \mathbf{P(n,k)}\)

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 both the cases of permutation in details.

(1) 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:

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

(2) 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,

\(\large n \times n \times n \times ……(r \;\;times)\)

which is given as \(\large n^{r}\)

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) = \(\large 5^{3}\)

\(\large = 125\)

(3) 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:

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

Example: 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 arrangement will be,

\( 5! \times 2! = 120 \times 2 = \) 240 ways

(ii) The total number of arrangement 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: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

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Practise This Question

The sum of the digits in the unit place of all numbers formed with the help of 3,4,5,6 taken all at a time is