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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 ?

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    $\begingroup$ Two C's together: just spell it as "SUCESS". $\endgroup$
    – drhab
    Mar 20 '15 at 8:31
  • $\begingroup$ For two C's together number of ways is 6!/2..what to do next? $\endgroup$
    – user220382
    Mar 20 '15 at 8:33
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    $\begingroup$ have a look here $\endgroup$
    – drhab
    Mar 20 '15 at 8:38
  • $\begingroup$ I haven't started real thinking about it yet, and must leave now. It's not unthinkable that later I come back, but this in the hope that you are ready with it at that time. Good luck. $\endgroup$
    – drhab
    Mar 20 '15 at 8:54
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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 $3!*4=24$.

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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 :

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

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    $\begingroup$ You need to add the case of $SCCS**S$ and its symmetric counterpart to get a total of 24 options. $\endgroup$
    – Cain
    Mar 20 '15 at 9:20
  • $\begingroup$ Yes, of course, thanks for improving my answer $\endgroup$ Mar 20 '15 at 9:20

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