Question

How many permutations can be made such that the vowels can occupy only even positions?

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

Answer 1

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

$\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$

$PERMUTATION$

Vowels $E,U,A,I,O⟶5$ vowels

Consonants $:P,R,T,T,M,N⟶6$ consonants

Where $2$ are repeated

$5$ places , $5$ vowels

Number of ways of arranging $=5!$

$6$ places , $6$ Consonants

Number of ways of arranging $=\frac{6!}{2}$

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