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You are here: AIEEE Classes » AIEEE Prep Questions » Math » Permutation Combination » Question 5

Permutation : Rearranging Letters of words

Question 5
The number of arrangements of the word SLEEPLESSNESS where no 2 Ss appear adjacent to each other is
  1. 13!/(4!*2!*5!)
  2. 13!/(4!*2!*3!)
  3. 8c5*8!/(4! * 2!)
  4. 9c5*8!/(4! * 2!)
Correct Answer is 9c5*8!/(4! * 2!). Choice (D).

Explanatory Answer

The quickest way of solving this problem in permutation would be to arrange all the letters in this word except the Ss, and then try and place the Ss in a way such that no two Ss are adjacent.

Apart from the 5 Ss there are 4 Es, 2 Ls, 1 P and 1 N.

These can be permuted in 8!/(4!*2!).

After we do this, we need to place the 5 Ss in such a way that no 2 Ss are adjacent.

Once we have these 8 slots taken up, there are 9 slots free for the Ss to go into.

i.e., 7 slots in between any two adjacent letters, 1 right in the beginning and 1 right in the end.

The 5 Ss have to be placed within these 9 slots. This can be done in 9C5 ways.

So, the right answer is 9c5*8!/(4! * 2!) or choice D.

This is actually one of my favourite questions. The way in which the "should not be adjacent" problem is tackled is beautiful. If one goes by the brute force method of finding the overall and then subtracting, it would take ages to solve this problem.



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