<s>[INST] Evaluate the sum $\sum_{n=1}^{b}n\binom{a}{n}\binom{b}{n}$ Let $a$ and $b$ be natural numbers such that $a \geq b \geq 1$. How can we evaluate the following sum?
$$ \sum_{n=1}^{b}n\binom{a}{n}\binom{b}{n}$$
 [/INST] For any polynomial $p(z) = a_0 + a_1 z + \cdots a_n z^n$, let $[z^k]p(z)$ be the
coefficient $a_k$ for the monomial $z^k$ in $p(z)$. We have
$$\begin{align}\sum_{n=1}^b n \binom{a}{n}\binom{b}{n}
&= \sum_{n=0}^b \binom{a}{a-n}\times n\binom{b}{n}
= \sum_{n=0}^b [z^{a-n}]\{(1+z)^a\} \times [z^n]\{z\frac{d}{dz}(1+z)^b\}\\
&= [z^a]\{(1+z)^a \times bz(1+z)^{b-1}\}
= b[z^{a-1}]\{(1+z)^{a+b-1}\}\\
&= b\binom{a+b-1}{a-1}
= \frac{(a+b-1)!}{(a-1)!(b-1)!}
\end{align}
$$
</s>