AMC 12A 2003 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?

Answer Choices
A. \sum_{k=0}^{5}\left(\begin{array}{l}5 \\ k\end{array}\right)^{3}
B. 3^{5} \cdot 2^{5}
C. 2^{15}
D. \dfrac{15 !}{(5 !)^{3}}
E. 3^{15}