Problem: For nonnegative integers $a$ and $b$ with $a+b \leq 6$, let $T(a, b)=\binom{6}{a}\binom{6}{b}\binom{6}{a+b}$. Find the sum of all $T(a, b)$, where $a$ and $b$ are nonnegative integers with $a+b \leq 6$. Solution: Notice that the answer is $ \sum_{b = 0}^6\sum_{a=0}^6{6 \choose b} {6 \choose a} {6 \choose a+b} = \sum_{b = 0}^6{6 \choose b}\sum_{a=0}^6{6 \choose a} {6 \choose a+b} $ because when $a + b > 6$, ${6 \choose a + b} = 0$. For a fixed $b$, if $X,Y$ are disjoint sets with $|X| = |Y| = 6$ then $\sum_{a = 0}^6{6 \choose a} {6 \choose a+b} = \sum_{a = 0}^6{6 \choose 6-a} {6 \choose a+b}$ counts the number of tuples $(A,B)$ such that $A \subseteq X$ and $B \subseteq Y$ where $|B| = a+b$. Another way to count this is to simply choose $6+b$ elements from $X \cup Y$; if the elements are $X$ then they get assigned to $A$ and if they are in $Y$ then they get assigned to $B$. Hence, $\sum_{a = 0}^6{6 \choose a} {6 \choose a+b} = \sum_{a = 0}^6{6 \choose 6-a} {6 \choose a+b} = {12 \choose 6+b}$ and so the sum becomes $\sum_{b = 0}^6{6 \choose b}{12 \choose 6+b} = \sum_{b = 0}^6{6 \choose b}{12 \choose 6-b} = \binom{18}{6}.$ Alternate: Let $c = 6-(a+b)$ so that $a+b+c = 6$. Now, the problem is identical to summing ${6 \choose a}{6 \choose b}{6 \choose c}$ so that $a+b+c = 6$ (because then $c = 6-(a+b)$). This is equivalent to selecting $6$ tiles from a $6 \times 3$ grid (where each column denotes the value of $a,b,c$) which is just ${18 \choose 6}$.