Question

Consider the word ${}^{\prime }PERMUTATIO{N}^{\prime }$. How many permutations can be made such that all the vowels and consonants are alternate?

• A. $\frac{7!}{2!}$
• B. $\frac{5!\times 6!}{2!}$
• C. $5!\times 6!$
• D. none of these

Answer 1

The correct option is B $\frac{5!\times 6!}{2!}$

$PERMUTATION$

$6$ Consonants $5$ Vowels

$6$ odd places $5$ even places

$\Rightarrow$ Number of ways of arranging alternately

$=\frac{6!}{2!}\times 5!=\frac{6!\times 5!}{2!}$