In how many ways can the letters of the word | vedclass

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

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(C) The word $EAMCET$ contains $6$ 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: $E, A, M, C, E, T$. Vowels: $E, A, E$. Consonants: $M, C, T$. Total letters = $6$. Vowels = $3$ $(E, A, E)$. Consonants = $3$ $(M, C, T)$. To ensure no two vowels are together,we arrange the consonants first: $3! = 6$ ways. This creates $4$ gaps: $\_ C \_ C \_ C \_$. We need to place $3$ vowels in these $4$ gaps. The number of ways to choose and arrange the vowels is $^4P_3 = 4 	imes 3 	imes 2 = 24$. Since there are two identical vowels $(E, E)$,we divide by $2!$. Total arrangements = $3! 	imes \frac{^4P_3}{2!} = 6 	imes \frac{24}{2} = 6 	imes 12 = 72$.

In how many ways can the letters of the word $EAMCET$ be arranged such that no two vowels are together?

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