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

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) (8 !/3 ! 3 !) (B) (5 !/3 !× 6C2) (C) (5 !/3 !)× 6C3 (D) (8 !/5 !)× 6C3

Tardigrade Admin · Accepted Answer

First place N, I, N, T, N at cross places X.X.X.X.X. Then place Eâs at dot places. In that case, no two Eâs will be together. ∴ total no. of ways =(5 !/3 !)⋅ 6C3 [∵ no. of Nâs = 3 out of 5 N, I, N, T, N]

Q. 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

$\frac{8 !}{3 ! 3 !}$

$\frac{5 !}{3 ! \times ^{6} C _{2}}$

$\frac{5 !}{3 !} \times^{6} C_{3}$

$\frac{8 !}{5 !} \times^{6} C_{3}$

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.
$∴$ total no. of ways $= \frac{5 !}{3 !} \cdot^{6} C_{3}$
[ $∵$ no. of $N ’ s = 3$ out of $5 N, I, N, T, N$ ]

Q. 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

3!3!8!​

3!×6C2​5!​

3!5!​×6C3​

5!8!​×6C3​