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