Question

The number of permutations that can be made out of the letters of the word "ENTRANCE" so that the two 'N's are always together is

• A. $\frac{8!}{(2!{)}^2}$
• B. $8!$
• C. $\frac{7!}{2!}$
• D. $\frac{7!}{(2!{)}^3}$

Answer 1

Answer: (C)

$\frac{7!}{2!}$

Total number of ways such that both N's are always together is nothing but considering both N's as single element and checking the number of permutations.
As the word has a repitition E twice
$\therefore$ Total number of ways are $\frac{7!}{2!}$.