Question

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

• A. $\frac{8\,!}{3\,!\,3\,!}$
• B. $\frac{5\,!}{3\,!\times\,{}^6C_2}$
• C. $\frac{5\,!}{3\,!}\times \,{}^6C_3$
• D. $\frac{8\,!}{5\,!}\times \,{}^6C_3$

Answer 1

Answer: (C)

First place $N, I, N, T, N$ at cross places
$X.X.X.X.X.$
Then place $E’s$ at dot places. In that case, no two $E’s$ will be together.
$\therefore $ total no. of ways $=\frac{5\, !}{3\,!}\cdot \,{}^{6}C_{3}$
[$\because$ no. of $N’s = 3$ out of $5 \,N, I, N, T, N$]