By permuting the letters of word "SUCCESS" the number of ways in which no 2 C's and no 2 S's are together are? 
By permuting the letters of word "SUCCESS" the number of ways in which
  no 2 C's and no 2 S's are together are?

I am getting is as $$\frac{7!}{3!2!}-(5!-4!) $$ using inclusion exclusion.Am I correct?
Update:5! is number of ways in which two S's and 2 C's are together.However there has been over-counting of the case where three S's and 2 C's exist.So I subtracted 4!.Now total number of possibilities of permutation is $7!/(2!*3!)$.Now I subtracted the unfavorable cases to get the above answer.
 A: With the inclusion-exclusion principle, I would like to compute the following:
(Total number of permutations) - (Number of permutations with $2$ $C$'s together) - (Number of permutations with $2$ $S$'s together) + (Number of permutations with $2$ $C$'s together and $2$ $S$'s together) + (Number of permutations with $3$ $S$'s together) - (Number of permutations with $2$ $C$'s together and $3$ $S$'s together).
Total number of permutations: $\frac{7!}{3!2!}$.  There are $7!$ orderings, but we divide out by the repeated letters.
Number of permutations with $2$ $C$'s: $\frac{6!}{3!}$.  There are $6!$ orderings when we treat $CC$ as a single character, but we divide out by the repeated $S$'s.
Number of permutations with $S$ $S$'s: $\frac{6!}{2!}$.  Put $2$ of the $S$'s together and leave the third one out of the pair.  Divide out by the repeated $C$'s.
Number of permutations with $2$ $C$'s and $2$ $S$'s: $5!$.  Put the $2$ $C$'s and $2$ of the $S$'s as a single letter.
Number of permutations with $3$ $S$'s: $\frac{5!}{2!}$.  Put all $3$ of the $S$'s together and divide out by the $C$'s.
Number of permutations with $2$ $C$'s and $3$ $S$'s: $4!$.  Put the $2$ $C$'s and $3$ $S$'s as single letters.
A: The wording "no $2 C's$ and no $2 S's$ are together" surely means that $C's$ and $S's$ are never allowed to be together ? I am interpreting it as thus,
 and applying the "gap method" followed by the "subtraction method".
Firstly, we shall keep the $S's$ separate by placing them in the gaps of $-U-C-C-E-$ and permute the other letters, thus $\binom53\cdot\frac{4!}{2!} = 120$ ways.
We shall now subtract arrangements with the $C's$ together treating them as a super $C$,
$ -U-\mathscr C - E - \;,$  thus $\binom43\cdot3! = 24$
thus the final ans is $120 -24 = 96$
