2003 AMC 12A Problem 20

Suppose the first block holds $k$ B’s and $5-k$ C’s. The remaining $k$ C’s must go in the second block (since the third has no C’s), forcing $5-k$ A’s there. Then the third block contains the remaining $k$ A’s and $5-k$ B’s. For each $k,$ the $k$ B’s in the first block, $k$ C’s in the second, and $k$ A’s in the third can be placed in $\binom{5}{k}^3$ ways, so the total is $\displaystyle\sum_{k=0}^{5}\binom{5}{k}^3.$ Thus, the correct answer is A.