Question

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

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

Answer 1

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

Given, word NINETEEN
No 3 E's should come together hence E's can be arranged in $\frac{{}^6{\text{P}}_3}{3!}$ ways.
And the remaining 5 letters can be arranged is $5!$ ways and as N is repeating trice divide by $3$! ways
$\therefore$ the total number of ways are $\frac{5!}{3!}\frac{{}^6{\text{P}}_3}{3!}$ ways
=$\frac{5!}{3!}{\times }^6{\text{C}}_3$ ways