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# 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,

nPr = (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,

nCr = (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:

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)

nCr = (n!)/[r! (n-r)!]

= 3Cr Therefore, there are three ways of choosing two items.

Now, in case of permutations,

AB

BA

BC

CB

CA

AC

nPr = (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,

nPr = (n!)/(n-r)!, n>0 and r>0

Here,

n = 6

r = 4

Substituting the values, we get,

nPr = (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,

nCr = (n!)/[r! (n-r)!], n>0 and r>0

Here,

n = 6

r = 4

Substituting the values, we get,

nPr = (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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