2017 AIME I 第 7 题

Replace $\binom{6}{a+b}$ by $\binom{6}{6-a-b},$ so with $c = 6 - a - b$ each term becomes $\binom{6}{a}\binom{6}{b}\binom{6}{c}$ with $a + b + c = 6$

Choosing $a,$ $b,$ and $c$ elements from three disjoint $6$-element sets, over all $a + b + c = 6,$ counts the $6$-element subsets of an $18$-element set

By the symmetry $\binom{6}{a+b} = \binom{6}{6-(a+b)},$ substituting $c = 6 - a - b$ turns the sum into $$S = \sum_{a+b+c=6} \binom{6}{a}\binom{6}{b}\binom{6}{c}.$$

Each term counts the ways to choose $a$ elements from one $6$-element set, $b$ from a second, and $c$ from a third. Summed over all $a + b + c = 6,$ this counts every way to choose $6$ elements from the combined $18$-element set, so $S = \binom{18}{6} = 18564.$

The remainder upon division by $1000$ is $564.$