AIEEE Discussion Forum, Questions

Question
In how many ways can the letters of the word SLEEPLESSNESS be rearranged such that no two Ss are adjacent?
a. 13!/(4!*2!*5!)
b. 13!/(4!*2!*3!)
c. 8c5 * 8!/(4!*2!)
d. 9c5 * 8!/(4!*2!)

13 letters in the word. 5 S, 4 E and 2 L.

So, the answer is 13! / (5! * 4! * 2!)

Is this correct?

Hi Balamurugan,

that would be the right answer if the constraint were not there. The answer 13! / (5! * 4! * 2!) would include all the combinations where the S's would appear together as well. From this we would have to substract the instances where all 5S's appear together, and then 4S's appear together etc.

The other way of going about this would be to arrange all the letters in this word except the S's, and then trying to place the S's in somew way where they are not adjacent.

Apart from the S's there are 4 E's. 2 L's, 1 P and 1 N, these can be permuted in 8!/(4!*2!). After we do this, we need to place the S's in such a way that no 2 S's are adjacent.

Once we have these 8 slots taken up, there are 9 slots free for the S's to go into. 7 slots in between any two adjacent letters, 1 right in the beginning and 1 right in the end. The 5 S's 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.

Sir. Thanks for your answer.

Will I get the same answer if I find out the total number of sequences and then subtract the cases where 2 Ss are together, 3 Ss are together, 4 Ss are together and finally 5 Ss are together?

Hi Balamurugan,

We can take that approach also. But the problem here would be that although we can eliminate the ones where all 5 are together easily, everything else is a little complicated. For instance, when you try to compute ones where 2 S's are together, how would you account for the ones in which 2 sets of 2 S's appear together. Also, when the 5 S's appear as a triplet + a duo, how would you account for that? Importantly, when you are subtracting the ones in which the S's appear in a group of 4, one will also have to avoid counting the one combo where all 5 S's appear together as well. I guess, although the brute force may be possible, it might get too complicated.

I think it might be an interesting exercise to try this out and see how one can go about calculating this the brute force way. I would go for ikt by computing the number of arrangements for wach of these combinations

5 S's appearing together
S's appearing as 4 in one bunch and 1 separately
3S's and 2S's together
3+1+1
2+1+1+1
2+2+1

Thanks sir.

One more doubt. We are using 9C5 instead of 9P5 because these 5 Ss are same. Is that correct?

Hi Balamurugan,

You are right. We are selecting 9c5 because we are trying to choose 5 slots out of the 9 gaps available. The S's need to be merely placed in these gaps. If the 5 letters we were selecting were different, we would have to fret about arranging them in some order as well.