The number of different words that can be for | vedclass

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Question

(A) The word "$INTERMEDIATE$" consists of $12$ letters: $I, N, T, E, R, M, E, D, I, A, T, E$.Vowels are: $I, E, E, I, A, E$ (total $6$ vowels: $3$ $E$

Vedclass · Accepted Answer

$\frac{6!}{2!} 	imes \frac{7!}{2!3!}$

Vedclass · Answer

(A) The word "$INTERMEDIATE$" consists of $12$ letters: $I, N, T, E, R, M, E, D, I, A, T, E$.Vowels are: $I, E, E, I, A, E$ (total $6$ vowels: $3$ $E$'s,$2$ $I$'s,$1$ $A$).Consonants are: $N, T, R, M, D, T$ (total $6$ consonants: $2$ $T$'s,$1$ $N, 1$ $R, 1$ $M, 1$ $D$).First,arrange the $6$ consonants. The number of ways is $\frac{6!}{2!}$.These $6$ consonants create $7$ gaps (including ends) where the $6$ vowels can be placed so that no two vowels are together.The number of ways to arrange the $6$ vowels in these $7$ gaps is $^7P_6$,but since there are repetitions among vowels ($3$ $E$'s and $2$ $I$'s),we divide by $3!2!$.Thus,the number of ways is $\frac{^7P_6}{3!2!} = \frac{7!}{3!2!}$.Total arrangements = $\frac{6!}{2!} 	imes \frac{7!}{3!2!}$.

The number of different words that can be formed from the letters of the word "$INTERMEDIATE$" such that two vowels never come together,is

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