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

Askiitians Expert Ankit Jain- IIT Bombay
18 Points
13 years ago

Its a nice question you have posted.

The word examination consists of letters in following frequencies.

E=1; X=1; A=2; M=1; N=2; T=1; O=1; I=2;

For picking up the 4 letters and arrange them into words woul require us to consider 3 cases.

CASE I:  2 same letters each of 2 kinds

THe possibilities are:

AANN

AAII

NNII

Each word can be arranged in 4!/2!*2! ways. So for three cases total ways = 3* 4!/2!*2! =18 ways.

CASE II: 2 same letters of one kind

The possbilities are:

AA _ _

NN _ _

II _ _

Now consider the case with AA _ _. For selecting the next two letter you can select maximum of 1  N and 1 I. So you need to select 2 letters from 7 left. ( Total =11. 2 left for AA , 1 N and 1 I left).

So say for AA  _ _ case total number of permutations = C(7,2) 4!/2!

So total permutations for all the three cases = 3 * C(7,2) 4!/2! = 378 ways

CASE III: No repetition of any letter

Total no of permutation for this can be directly written as = C(8,4) * 4! =1680 ways

So total number of ways = 1680 + 378 + 18 =2076 ways

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