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

6 years ago

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Answers : (1)

										

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!!


Regards

Askiitians Expert

Ankit Jain- IIT Bombay

6 years ago

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