2017
Problem - 3786
For nonnegative integers $a$ and $b$ with $a + b \leq 6$, let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$. If $S$ denotes the sum of all $T(a, b)$, where $a$ and $b$ are nonnegative integers with $a + b \leq 6$. Find $S$.
Let $c=6-(a+b)$. Then we have $$\binom{6}{a+b}=\binom{6}{c} \implies T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{c}$$
Then $S$ equals sum of $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ when $a$, $b$, and $c$ are non-negative integers which satisfy the relation $a+b+c=6$.
Employing the generalized Vandemonde's identity (# 4275) yields $$\sum_{a+b+c=6}\binom{6}{a} \binom{6}{b} \binom{6}{c}=\binom{6+6+6}{a+b+c}=\binom{18}{6}=\boxed{18564}$$