Question

Find the number of ways in which the letters of the word ${}^{\prime }AEROPLAN{E}^{\prime }$ can be arranged such that the vowels are always together.

• A. $\frac{5!}{2!}$
• B. $\frac{5!}{2{!}^2}$
• C. $\frac{5{!}^2}{2{!}^2}$
• D. $\frac{9!}{2{!}^2}$

Answer 1

The correct option is C $\frac{5{!}^2}{2{!}^2}$

${}^{\prime }AEROPLAN{E}^{\prime }$ Total number of letters $=9$
Vowels - $2A,2E,1O$
Considering all vowels as a single letter, we are left with $5$ letters
They can be arranged in $5!$ ways
Now vowels can be arranged among themselves in $\frac{5!}{2!.2!}$
So required number of ways $=\frac{5{!}^2}{2{!}^2}=3600$