2003 AMC 12A Problem 20

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}

In the first block (no A’s), suppose there are

k

B’s and

5-k

C’s

Fixing

k

determines the letter counts in all three blocks, each chosen with

\binom{5}{k}

placements

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.

Problem 20 in other years