All the letters of the word EAMCET are arrang | vedclass

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Question

(C) The word $EAMCET$ contains $6$ letters: $E, A, M, C, E, T$.There are $3$ consonants: $M, C, T$ and $3$ vowels: $E, A, E$.First,we arrange the $3$

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(C) The word $EAMCET$ contains $6$ letters: $E, A, M, C, E, T$.There are $3$ consonants: $M, C, T$ and $3$ vowels: $E, A, E$.First,we arrange the $3$ consonants in $3!$ ways.$3! = 3 	imes 2 	imes 1 = 6$ ways.To ensure no two vowels are adjacent,we use the gap method. There are $4$ possible gaps around the $3$ consonants: $\_ C_1 \_ C_2 \_ C_3 \_$.We need to place $3$ vowels $(E, A, E)$ in these $4$ gaps. The number of ways to choose $3$ gaps out of $4$ is $^4C_3 = 4$.Since there are two identical vowels $(E, E)$,the number of arrangements of the vowels in the chosen gaps is $\frac{3!}{2!} = 3$.Total arrangements $= 3! 	imes ^4C_3 	imes \frac{3!}{2!} = 6 	imes 4 	imes 3 = 72$.

All the letters of the word $EAMCET$ are arranged in all possible ways. The number of such arrangements in which two vowels are not adjacent to each other is

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