Combinations And Permutations Of A Word

Question

Different words are being formed by arranging the letters of the word "SUCCESS". Number of words in which two C's are together but no two S's are together ?

Answer 1

Treat the two Cs as a single letter. Then arrange all non S letters, there are 3 such letters (the Cs are counted as one). Over all $3!$ arrangements.

Now, to make sure no two S are adjacent it'll suffice to put at most one S between any of the existing letters (including the extremities). Since there are 3 letters there are 4 possible locations, out of which 3 locations need to be chosen, total of 4 possibilities. Overall the number of arrangements is $3!*4=24$.

Answer 2

You have 7 places to put your letters 1 to 7. In my answer the symbol $*$ means "still not filled place".

First choose the place where you are going to put the two C's. Since they must be together you have got 6 choices :

$$CC*****,*CC****,**CC***,***CC**,****CC*,*****CC$$

One can easily see that, doing a symmetry, the last three cases correspond bijectively to the first three cases. So we now just work on the first three cases.

Now for each choice of a place for CC you have got to put three S's not next to each other. You then easily get the following cases :

$$CCS*S*S,SCCS*S*,SCC*S*S,SCCS**S,S*CCS*S,*SCCS*S $$

And of course the symmetric ones :

$$S*S*SCC,*S*SCCS,S*S*CCS,S**SCC,S*SCC*S,S*SCCS* $$

Finally for each such choice of places for the C's and the S's you have only two choices left : namely, how to place the remaining letter $U,E$. So that you have on the whole :

$$12\times 2=24\text{ choices.}$$