# success arrangement

• August 20th 2011, 09:45 AM
success arrangement
in how many letters of SUCCESS can be arranged such that no two C or no two S are together.
• August 20th 2011, 10:13 AM
Plato
Re: success arrangement
Quote:

Let $\mathcal{S}$ be the set of all rearrangements in which no to S's are together.
Let $\mathcal{C}$ be the set of all rearrangements in which no to C's are together.
The number in $\mathcal{C}$ is $\|\mathcal{C}\|=\binom{6}{2}\frac{5!}{3!}$.
Now you must calculate $\|\mathcal{C}\cup\mathcal{S}\|.$