Permutation vs Combination – Definition, Differences, Examples

Permutation is referred to as the selection followed by an arrangement of a certain set of items from a given sequence or collection of items. Permutations are denoted by the following formula, 

ⁿP_r = (n!)/(n-r)!, n > 0 and r > 0

where,

n is considered to be the number of different elements.

r is the arrangement pattern of the element.

Combination 

Combination is referred to as the selection of a certain set of items from a given sequence or collection of items. The order of selection of individual items doesn’t hold any relevance in the case of combinations. Combinations are denoted by the following formula, 

ⁿC_r = (n!)/[r! (n-r)!], n>0 and r>0

where 

n is the number of different elements

r is the arrangement pattern of the element

The key differences between permutation and combination are as follows: 

Permutation	Combination
Ordered arrangement of items	Unordered arrangement of items
Also known as ordered elements	Also known as an unordered set of elements
A single combination may lead to the derivation of multiple permutations.	A single permutation can lead to only a single combination.
Order is important.	Order does not hold any relevance.

To illustrate the difference between permutation and combination we can consider the following examples; 

Let us suppose we have 3 alphabets, A, B, and C. 

Now, 

Possible combinations of selection of any two items from three are as follows: 

AB  (=BC)

BC  (=CB)

CA  (=AC)

ⁿC_r = (n!)/[r! (n-r)!]

= ³C_r

Therefore, there are three ways of choosing two items. 

Now, in case of permutations, 

AB  

BA

BC

CB

CA  

AC 

ⁿP_r = (n!)/(n-r)!

Therefore, there are six ways of choosing and arranging two items from a total of six items. 

Sample Questions

Question 1: Give an example of permutation and combination of two items Yash and Yashvi 

Solution: 

Possible Permutations are

Yash Yashvi 

Yashvi Yash

Possible Combinations are

Yash Yashvi, since both the items have to be definitely selected

Question 2: Calculate the permutation of selecting 4 items from 6 items.

Solution: 

We know, 

ⁿP_r = (n!)/(n-r)!, n>0 and r>0

Here, 

n = 6

r = 4

Substituting the values, we get, 

ⁿP_r = (6!)/(6-4)!

= 6!/2!

= 6 * 5 * 4 * 3

= 360

Question 3: Calculate the combination of selecting 4 items from 6 items.

Solution: 

We know, 

ⁿC_r = (n!)/[r! (n-r)!], n>0 and r>0

Here,

n = 6

r = 4

Substituting the values, we get, 

ⁿP_r = (6!)/[4! * (6-4)!]

= 6! / (4! * 2!)

= 6!/ 4! * 2

= (6 * 5 * 4 * 3 * 2 )/ (4 * 3 * 2) * 2

= 15

Question 4: Calculate 5!, if suppose, 4! = 100

Solution: 

We know, 

n! = n * (n-1)!

Substituting the given hypothetical values, we get, 

5! = 5 * 4!

= 5 * 4!

= 5 * 100

= 500

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