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(C) The word $EAMCET$ has $6$ letters: $E, A, M, C, E, T$.There are $2$ vowels $(E, E)$ and $4$ consonants $(A, M, C, T)$. Wait,let us re-examine the

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(C) The word $EAMCET$ has $6$ letters: $E, A, M, C, E, T$.There are $2$ vowels $(E, E)$ and $4$ consonants $(A, M, C, T)$. Wait,let us re-examine the letters: $E, A, M, C, E, T$. The vowels are $E, A, E$ ($3$ vowels) and consonants are $M, C, T$ ($3$ consonants).First,arrange the $3$ consonants $M, C, T$ in $3! = 6$ ways.These consonants create $4$ gaps (including ends): $\_ M \_ C \_ T \_$.We need to place the $3$ vowels $(E, A, E)$ in these $4$ gaps such that no two vowels are together.The number of ways to choose $3$ gaps out of $4$ is $^4C_3 = 4$.Since there are $2$ identical vowels $(E, E)$,the number of ways to arrange the vowels in the chosen gaps is $\frac{3!}{2!} = 3$.Total arrangements = $3! 	imes ^4C_3 	imes \frac{3!}{2!} = 6 	imes 4 	imes 3 = 72$.

Find the total number of arrangements of the letters of the word $EAMCET$ such that no two vowels are together.

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