combinatorics - Sum of Some Binomial Terms Equals Zero

Let $q$ and $\ell$ be positive integers. Then the sum
$$
\sum_{k=q}^\ell (-1)^{k+q}\binom{k}{q}\binom{\ell}{k} =
\left\{\begin{array}{ccc} 1 \mbox{ if } \ell =q\\ 0 \mbox{ if }\ell >q\end{array}\right.
$$

I don't see what to do here. Can we interpret it combinatorially somehow?

Answer

HINT: Assume that $\ell>q$, since the other case is trivial. Suppose that we have white slips of paper numbered $1$ through $\ell$. If $k\le q\le\ell$, $\binom{\ell}k\binom{k}q$ is the number of ways to choose $k$ of those slips and color them blue, then choose $q$ of the $k$ blue slips and circle the number. This is clearly the same as choosing $q$ of the original $\ell$ slips, coloring them blue, and circling the number, and then choosing $k-q$ of the remaining $\ell-q$ white slips and coloring them blue, so $\binom{\ell}k\binom{k}q=\binom{\ell}q\binom{\ell-q}{k-q}$. And $k+q\equiv k-q\pmod2$, so, setting $j=k-q$,

$$\sum_{k=q}^\ell(-1)^{k+q}\binom{\ell}k\binom{k}q=\binom{\ell}q\sum_{j=0}^{\ell-q}(-1)^j\binom{\ell-q}j\;.$$

Can you finish it from there?