Question

All the letters of the word $EAMCET$ are arranged in all possible ways. The number of such arrangements in which no two vowels are adjacent to each other is

• A. $36$
• B. $54$
• C. $72$
• D. $144$

Answer 1

The correct option is C $72$

Vowels are $E,A,E$ and consonants are $M,C,T$

For two vowels not to come adjacent, first of all arrange the $3$ consonants in $3!$ ways in a line leaving the adjacent sides vacant.

Now these $3$ consonants will have $4$ vacant spaces, and we have $3$ vowels

So we can select $3$ vacant spaces for vowel in ${}^4{C}_3$ ways and then arrange them in $\frac{3!}{2!}$ ways. ( since $E$ is repeated )

Hence required number of word will be $=3!{\cdot }^4{C}_3\cdot \frac{3!}{2!}=72$