Question

The number of permutations that can be made from the letters of the word "HOTEL" so that the vowels may occupy the even places is

• A. $2$
• B. $6$
• C. $12$
• D. $36$

Answer 1

Answer: (C)

$12$

In the word $HOTEL$ vowels are $O$ and $E$

Since, these two letters can occupy the second or fourth place, these can be done in $2!$ ways.

The remaining three letters $H$, $T$ and $L$ can occupy any of the first, third or fifth place.

These can be done in $3!$ ways.

$\therefore$ the total no. of permutations is $2!\times 3!=12$