AMC 12 2003a Problem 20

How many $15$-letter arrangements of $5$ A's, $5$ B's, and $5$ C's have no A's in the first $5$ letters, no B's in the next $5$ letters, and no C's in the last $5$ letters?

$\textrm{(A)}\ \sum_{k=0}^{5}\binom{5}{k}^{3} \textrm{(B)}\ 3^{5}\cdot 2^{5} \textrm{(C)}\ 2^{15} \textrm{(D)}\ \frac{15!}{(5!)^{3}} \textrm{(E)}\ 3^{15}$