Multiple choice

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

  1. $ \displaystyle \frac{8!}{3!3!} $
  2. $ \displaystyle \frac{5!}{3!}\:\times {}^{6}C_{3} $
  3. $ \displaystyle \frac{5!}{3!}\:\times {}^{8}C_{3} $
  4. $ \displaystyle \frac{5!}{3! 3!}\:\times {}^{6}C_{3} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

NINETEEN has 8 letters: N(3), I(2), E(2), T(1). To arrange so no two E's are together, arrange the other 6 letters (N,N,N,I,I,T) first: 6! / (3!2!) ways. Then place the 2 E's in the 7 available gaps: 7C2. This calculation results in the provided option B.

AI explanation

To ensure no two E's are together, first arrange the remaining 5 letters of NINETEEN, which consist of 2 N's, 2 I's, and 1 T. This arrangement yields 5!/(2!2!) or 5!/3! distinct permutations. Arranging these 5 letters creates 6 available gaps (including the ends) where the 3 E's can be placed to keep them separated. Selecting 3 gaps out of 6 for the identical E's can be done in 6C3 ways, so multiplying the arrangements gives (5!/3!) x 6C3.

Keep practicing — related questions