Question

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?

$\displaystyle\sum_{k=0}^{5}\binom{5}{k}^3$

$3^5 \cdot 2^5$

$2^{15}$

$\dfrac{15!}{(5!)^3}$

$3^{15}$

Solution:

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.

2003 AMC 12A Problem 20

Solution:

Problem 20 in Other Years