# Find the number of combinations and permutations of 4 letters taken from the word ‘EXAMINATION.’

Given the word is EXAMINATION. The letters present in it are:

E: 1 in number

X: 1 in number

A: 2 in number

M: 1 in number

I: 2 in number

N: 2 in number

T: 1 in number

O: I in number

a. We need to find the no. of words formed by 4 letters from the word EXAMINATION:

The possible cases are the following:

i. 4 distinct letters

ii. 2 alike letters and 2 distinct letters.

iii. 2 alike letters of one type and 2 alike letters of another type

i. There are 8 different letters from which we need to select 4 letters. Let us assume no. of ways of selection be N1

N1 = no. of ways of selecting 4 letters from 8 letters

N1 = 8C4

We know that ,

And also n! = (n)(n – 1)......2.1

N1 = 70

Now we need to arrange the chosen 4 different letters. Since every word differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of arrangements that can be made is 4!.

Let us assume no. of ways of arrangement be N2.

N2 = N1 × 4!

N2 = 70 × 24

N2 = 1680

ii. There are 3 letters which occurred more than once. So, we need to select 1 letter from these 2 and 2 distinct letters from the remaining 7 distinct letters. Let us assume no. of ways of selection be N3

N3 = (no. of ways of selecting 2 alike letters from the 3 types of alike letters) × (no. of ways of selecting 2 distinct letters from remaining 5 distinct letters)

N3 = (3C1) × (7C2)

We know that ,

And also n! = (n)(n – 1)......2.1

N3 = 3 × 21

N3 = 63

Now we need to arrange the chosen 2 different letters and 2 alike letters. The arrangement is similar to that of arranging n people in n places in which r are similar which are ways to arrange. So, the total no. of arrangements that can be made are .

Let us assume no. of ways of arrangement be N4.

N4 = N3 ×

N4 = 63 × 4 × 3

N4 = 756

iii. There are 3 letters which occurred more than once from which we need to select 2. Let us assume no. of ways of selection be N5

N5 = no. of ways of selecting 2 alike letters of one type and 2 alike letters of other type

N5 = 3C2

We know that ,

And also n! = (n)(n – 1)......2.1

N5 = 3

Now we need to arrange the chosen 2 alike letters of one type and 2 alike letters of another type. The arrangement is similar to that of arranging n people in n places in which r are similar of one type and m are similar of another type which are ways too arrange. So, the total no. of arrangements that can be made are .

Let us assume no. of ways of arrangement be N6.

N6 = N5 ×

N6 = 18

Total no. of combinations = N1 + N3 + N5

Total no. of combinations = 70 + 63 + 3

Total no. of combinations = 136

Total no. of ways of permutations = N2 + N4 + N6

Total no. of ways of permutations = 1680 + 756 + 18

Total no. of ways of permutations = 2454

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