Question

Find the number of words formed by permuting all the letters of the following words:
(i) INDEPENDENCE
(ii) INTERMEDIATE
(iii) ARRANGE
(iv) INDIA
(v) PAKISTAN
(vi) RUSSIA
(vii) SERIES
(viii) EXERCISES
(ix) CONSTANTINOPLE

Answer 1

There are 12 letters in the word `INDEPENDENCE' out of which 2 are D'S, 3 are N'S, 4 are E'S and the rest are all distinct.
So, the total number of words = $\frac{12!}{2!3!4!}$
= $\frac{12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4!}{2!3!4!}$
= $\frac{12\times 11\times 10\times 9\times 8\times 8\times 7\times 6\times 5}{2\times 3\times 2}$
= $11\times 10\times 9\times 8\times 7\times 6\times 5$
= 1663200
(ii) There are 12 letters in the word 'INTERMEDIATE' out of which 2 are I'S, 2 are T'S, 3 are E'S and the rest are all distinct.
So, the total number of words
= $\frac{12!}{2!2!3!}$
=$\frac{12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2!}{2!2!3!}$
= $\frac{12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3}{2\times 3\times 2}$
= $11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3$
= 19958400
(iii) There are 7 letters in the word 'ARRANGE'out of which 2 are A'S, 2 are R'S, and the rest are all distinct.
So, the total number of words = $\frac{7!}{2!2!}$
= $\frac{7\times 6\times 5\times 4\times 3\times 2!}{2!2!}$
= $\frac{7\times 6\times 5times4\times 3}{2\times 1}$
=1260
(iv) There are 5 letters in the word 'INDIA' out of which 2 are I'S, and the rest are all distinct.
So, the total number of words = $\frac{5!}{2!}$
= $\frac{5\times 4\times 3\times 2}{2!}$
= 60
(v) There are 8 letters in the word 'PAKISTAN' out of which 2 are A'S, and the rest are all distinct.
So, the total number of words
=$\frac{8!}{2!}$
= $\frac{8\times 7\times 6\times 5\times 4\times 3\times 2!}{2!}$
= $8\times 7\times 6\times 5\times 4\times 3$ = 20160
(vi) There are 6 letters in the rest are all distinct.
So, the total number of words
= $\frac{6!}{2!}$
= $\frac{6\times 5\times 4\times 3\times 2!}{2!}$
= $6\times 5\times 4\times 3=360$
(vii) There are 6 letters in the word 'SERIES' out of which 2 are S's, 2 are E's and the rest are all distinct.
So, the total number of words
= $\frac{6!}{2!2!}$
= $\frac{6\times 5\times 4\times 3\times 2!}{2!2!}$
= $\frac{6\times 5\times 4\times 3}{2\times 1}$
= $6\times 5\times 2\times 3=180$
(viii) There are 9 letters in the word 'EXERCISES' out of which 3 are E's, 2 are S's and the rest are all distinct.
So, the total number of words
words = $\frac{14!}{2!3!2!}$
= $\frac{14!}{2\times 3\times 2\times 2}$
= $\frac{14!}{2\times 3\times 2\times 2\times 2}$
= $\frac{14!}{24}$