JEE Class main Answered

The word contain 11 letters with repetition of letters.

We can choose 4 letters by

i) all the four letters are distinct

ii) Two distinct and two alike

iii) Two alike of one kind and two alike of other kind

i) 8 different letters are C, O, M, B, I, N, A, T

This can be chosen in ⁸C₄ × 4! = 1680 ways

ii) Three pairs alike letters OO, II, NN.

One pair can be chosen in ³C₁ = 3 ways 

Remaining two letters can be chosen by ⁷C₂ 

Each such group has 4 letters, out of which 2 are alike.

Hence, it can be chosen by 4!/2! 

Total number of ways = 3 × ⁷C₂ × 4!/2! = 756

iii) Three pairs of 2 alike letters. Two pairs can be chosen by ³C₂

 ³C₂ groups of 4 letters each.

There are 4 letters of which 2 alike of one kind and two alike of other kind in each group.

That can be arranged by 4!/2! × 1/2!

Total number of ways = ³C₂ × 4!/2! × 1/2! = 18

Hence, the total number of ways = 1680 + 756 + 18 = 2454