The number of arrangements of the letters of | vedclass

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(A) The word $INDEPENDENCE$ has $12$ letters: $I, N, D, E, P, E, N, D, E, N, C, E$.The vowels are $I, E, E, E, E$ (total $5$ vowels).The consonants ar

Vedclass · Accepted Answer

$16800$

Vedclass · Answer

(A) The word $INDEPENDENCE$ has $12$ letters: $I, N, D, E, P, E, N, D, E, N, C, E$.The vowels are $I, E, E, E, E$ (total $5$ vowels).The consonants are $N, N, N, D, D, P, C$ (total $7$ consonants).Since all vowels must occur together,we treat the group $(IEEEE)$ as a single entity.Now,we have $7$ consonants + $1$ vowel group = $8$ entities to arrange.The number of ways to arrange these $8$ entities,where $N$ repeats $3$ times and $D$ repeats $2$ times,is $\frac{8!}{3!2!}$.Within the vowel group $(IEEEE)$,the $4$ $E$'s are identical. The number of ways to arrange these $5$ vowels is $\frac{5!}{4!} = 5$.Total arrangements = $\frac{8!}{3!2!} 	imes \frac{5!}{4!} = \frac{40320}{6 	imes 2} 	imes 5 = 3360 	imes 5 = 16800$.

The number of arrangements of the letters of the word $INDEPENDENCE$ in which all the vowels always occur together is

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