Permutations - Formula & Examples
Permutation is the arranging of a given number of things in every possible order of succession. There is a particular rule to write the numbers, such as form the series of numbers 1, 2, 3, 4 etc, up to the number o things to be permuted, and their continued product will be the number of permutations.

For Example: How many different integral numbers may be expressed by writing the 5 significant digits in succession, each figure to be taken once, and once in each number. Its answer is 120 ( = 120).

The order or number of arrangements is known as the permutation. Basically, it is a rearrangement within elements of a given ordered list in one-to-one manner of correspondence. In this section will discuss in detail about permutation.


If we have to find permutation of a list of n elements, it is given by n!, read as n factorial. Also, n! can be calculated by finding the product of natural numbers from 1 to n, that is, 

n! = n (n – 1) (n – 2) ….. 3 . 2 . 1

For Example: we have three elements in a set say {4, 3, 7} so we have 3! = 6

arrangements which are {3, 4, 7}, {3, 7, 4}, {4, 3, 7}, {4, 7, 3}, {7, 4, 3}, {7, 3, 4}.

If we need to find permutation arrangement of k elements out of a given set of n elements then we have the following formula:

n (n – 1) (n – 2) … (n – k + 1)

It is denoted by various notations. P (n, k),  $nP_k$ etc.
The basic formula is given as follows:

$n P_k$ = $\frac{n!}{(n – k)!}$

This formula is equal to zero if $k > n$.

This formula gives the same product as given above. But it is just more simplified one.
But what is the difference between a combination and a permutation? Being precise we can say that if order of things does not matter then we say it is combination and if the order of things makes a difference then we say there is permutation.

For Example: the fruit salad is a combination but the combination of any safe will be a permutation.

So we say that a permutation is nothing but an ordered combination.

In permutation we have two types: one in which repetition is allowed and another without any repetition. 

For non repeating permutations we use the formula mentioned above either the factorial form or the k out of n elements form. 

For repeating ones we simply multiply n with itself the number of times it is repeating, that is, n^r where n is the number of things to be chosen from and r is the number of items being chosen.


Let us see some examples of problems based on permutations.
Example 1:  Find out the number of ways in which 4 girls can form a group of 15 in total to be lined up for a click.

Solution: We need to find out P(15,4)

P (15, 4) = $\frac{15!} {(15 - 4)!}$ = $\frac{15!}{11!}$ = $\frac{15 . 14 . 13 . 12 . 11!}{11!}$ 

 = 32760
So the group of 15 girls can form 32760 different lineups as 4.

Example 2: Given a word “MATHS”, find the number of possible arrangements of its letters.

Solution: We can see that there is no repetition in the word MATHS. Also, that there are 5 digits in total in the given word. Also, we have 5 out of 5 choices given.

Hence, P (5, 5) = $\frac{5!} {(5 – 5)!}$ = $\frac{5!}{0!}$ = 5! = 5 * 4 * 3 * 2 * 1 = 120.
So we have 120 numbers of possible arrangements of digits of word MATHS.

Example 3: Sujen wants to draw two cards from a deck of 52 cards, without replacement.  What is the probability of drawing a 10 and a jack in that order.

Solution: Sujen wants to draw 10 and a jack from a deck.

In a deck, there are four 10 and four jack 

So probability of drawing a 10 and a jack is $P(4, 1) . P(4, 1)$

From a deck of 52 cards Sujen wants to draw two cards i.e. P(52, 2) 

=> $\frac{ P(4, 1) .P(4, 1)} {P(52, 2)}$ = $\frac{ 4 . 4} {2652}$ = $\frac{ 16} {2652}$ = $\frac{4}{663}$

The probability of drawing a 10 and a jack is $\frac{4}{663}$.