in how many letters of SUCCESS can be arranged such that no two C or no two S are together.
Let $\displaystyle \mathcal{S}$ be the set of all rearrangements in which no to S's are together.
Let $\displaystyle \mathcal{C}$ be the set of all rearrangements in which no to C's are together.
The number in $\displaystyle \mathcal{C}$ is $\displaystyle \|\mathcal{C}\|=\binom{6}{2}\frac{5!}{3!}$.
Now you must calculate $\displaystyle \|\mathcal{C}\cup\mathcal{S}\|.$