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Hello Aditya,
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
Hope this solves your query.
Please feel free to post as many doubts on our disucssion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
All the best!!RegardsAskiitians ExpertAnkit Jain- IIT Bombay
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