Question

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

• A. $360$
• B. $144$
• C. $72$
• D. $54$

Answer 1

Answer: (B)

We note that there are $3$ consonants and $3$ vowels $E,A$ and $O$. Since no two vowels have to be together, the possible choice for vowels are the places marked as ${}^{\prime }{X}^{\prime }$ in $XMXCXTX$, these vowels can be arranged in ${}^4{P}_3$ ways, $3$ consonants can be arranged in $\lfloor 3$ ways. Hence, the required number of ways $=3!{\times }^4{P}_3=3!\times 4!=144$.