in how many letters of SUCCESS can be arranged such that no two C or no two S are together.

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- Aug 20th 2011, 09:45 AMayushdadhwalsuccess arrangement
in how many letters of SUCCESS can be arranged such that no two C or no two S are together.

- Aug 20th 2011, 10:13 AMPlatoRe: success arrangement
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 $\displaystyle \|\mathcal{C}\cup\mathcal{S}\|.$