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

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


2 Answers

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

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:




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!!


Askiitians Expert

Ankit Jain- IIT Bombay

Devraj konwar
11 Points
3 years ago
how many combination and permutations can be made with the letters of the word PARABOCA, taken three at a time?

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