Question

The number of permutations that can be formed by arranging all the letters of word $NINETEEN$ in which no two $E$ occur together is

• A. $\frac{8}{3!3!}$
• B. $\frac{5!}{3!{\times }^6{C}_2}$
• C. $\frac{5!}{3!}{\times }^6{C}_3$
• D. $\frac{8!}{3!}{\times }^6{C}_3$

Answer 1

The correct option is C $\frac{5!}{3!}{\times }^6{C}_3$

First, place $N,I,N,T,N$ at cross places

$.x.x.x.x.x.$

Then place $E^{\prime }s$ at dot places.

In that case, no two $E^{\prime }s$ will be together.

$\therefore$ Total number of ways $=\frac{5!}{3!}{.}^6{C}_3$

$[\because$ number of $N^{\prime }s=3$ out of $5(N,I,N,T,N)]$