Question

The number of permutations of the letters of the word EXERCISES taken 5 at a time is
A. 2250
B. 30240
C. 22660
D. None of these

Answer 1

Hint: We break the whole choices in different parts and evaluate them individually. We fix any one unique term out of the terms which are in the word in repetitions. We vary the other combinations for the rest of the letters.

Complete step by step solution:
We have the word ‘EXERCISES’. We have in total 9 letters out of which three E’s, two S’s and one each of X, R, C, I.
We have to find the number of permutations of the letters of the word EXERCISES taken 5 at a time.
There can be many types of arrangements or choices. We take one at a time and find the number.
We take three E’s, two S’s. Number of permutations will be
 $ \dfrac{5!}{2!\times 3!}=10 $ .
We take three E’s, two from the set of S, X, R, C, I. Number of permutations will be
$ {}^{5}{{C}_{2}}\times \dfrac{5!}{3!}=200 $ .
We take two E’s, two S’s and one from the set of X, R, C, I. Number of permutations will be
$ 4\times \dfrac{5!}{2!\times 2!}=120 $ .
We take two E’s, three from the set of S, X, R, C, I. Number of permutations will be
$ {}^{5}{{C}_{3}}\times \dfrac{5!}{2!}=600 $ .
We take one E, two S’s and two from the set of X, R, C, I. Number of permutations will be
$ {}^{4}{{C}_{2}}\times \dfrac{5!}{2!}=360 $ .
We take five from the set of E, S, X, R, C, I. Number of permutations will be
$ {}^{6}{{P}_{5}}=6!=720 $ .
We take two S’s and three from the set of X, R, C, I. Number of permutations will be
$ {}^{4}{{C}_{3}}\times \dfrac{5!}{2!}=240 $ .
The total number of permutations will be $ 10+200+120+600+360+720+240=2250 $ .
So, the correct answer is “2250”.

Note: We need to be careful about fixing any one side of having a variable number of E’s. we can do the same thing with S’s. we also need to be careful about repeatedly taking any choice twice.