Question 133
Compute
$$
\sum_{\substack{a+b+c=12 \\ a \geq 6,\ b,\ c \geq 0}} \frac{a!}{b!c!(a-b-c)!}
$$
where the sum runs over all triples of nonnegative integers $(a, b, c)$ such that $a+b+c=12$ and $a \geq 6$.
$$
\sum_{\substack{a+b+c=12 \\ a \geq 6,\ b,\ c \geq 0}} \frac{a!}{b!c!(a-b-c)!}
$$
where the sum runs over all triples of nonnegative integers $(a, b, c)$ such that $a+b+c=12$ and $a \geq 6$.