Question

Consider the word ${}^{\prime }PERMUTATIO{N}^{\prime }$. How many permutations can be made such that the vowels can occupy only odd places?

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

Answer 1

The correct option is C $\frac{6{!}^2}{2}$

$PERMUTATION$

$6$ places $⟶5$ vowels

${}^6{P}_5⟶$ Ways of arranging

Left places $6⟶6$ consonants

$\frac{6!}{2}⟶$ ways of arranging

Total ways

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

$=\frac{{(6!)}^2}{2}$ Ways