A Binomial Coefficient Sum: $\sum_{m = 0}^{n} (-1)^{n-m} \binom{n}{m} \binom{m-1}{l}$ In my work on $f$-vectors in polytopes, I ran across an interesting sum which has resisted all attempts of algebraic simplification. Does the following binomial coefficient sum simplify?
\begin{align}
\sum_{m = 0}^{n} (-1)^{n-m} \binom{n}{m} \binom{m-1}{l} \qquad l \geq 0
\end{align}
Update: After some numerical work, I believe a binomial sum orthogonality identity is at work here because I see only $\pm 1$ and zeros. Any help would certainly be appreciated.
I take $\binom{-1}{l} = (-1)^{l}$, $\binom{m-1}{l} = 0$ for $0 < m < l$ and the standard definition otherwise. 
Thanks!
 A: This is a special case of the identity $$\sum_k \binom{l}{m+k} \binom{s+k}{n} (-1)^k = (-1)^{l+m} \binom{s-m}{n-l},$$ which is identity 5.24 on p. 169 of Concrete Mathematics, 2nd edition.  With $l = n$, $m = 0$, $s = -1$, $k = m$, and $n = l$, we see that the OP's sum is $$(-1)^{2n} \binom{-1}{l-n} = \binom{-1}{l-n}.$$
This is $(-1)^{l-n}$ when $l \geq n$ and $0$ when $l < n$, as in Fabian's comment to Plop's answer.
A: $$\sum_{m=0}^n (-1)^{n-m} \binom{n}{m} \binom{m-1}{l} = (-1)^{l+n} + \sum_{l+1 \leq m \leq n} (-1)^{n-m} \binom{n}{m} \binom{m-1}{l}$$
So we need to compute this last sum. It is clearly zero if $l \geq n$, so we assume $l < n$.
It is equal to $f(1)$ where $f(x)= \sum_{l+1 \leq m \leq n} (-1)^{n-m} \binom{n}{m} \binom{m-1}{l} x^{m-1-l}$.
We have that $$\begin{eqnarray*} f(x) & = & \frac{1}{l!} \frac{d^l}{dx^l} \left( \sum_{l+1 \leq m \leq n} (-1)^{n-m} \binom{n}{m} x^{m-1} \right) \\
& = & \frac{1}{l!} \frac{d^l}{dx^l} \left( \frac{(-1)^{n+1}}{x} + \sum_{0 \leq m \leq n} (-1)^{n+1} \binom{n}{m} (-x)^{m-1} \right) \\
& = & \frac{1}{l!} \frac{d^l}{dx^l} \left( \frac{(-1)^{n+1}}{x} + \frac{(x-1)^n}{x} \right) \\
& = & \frac{(-1)^{n+1+l}}{x^{l+1}} + \frac{1}{l!} \sum_{k=0}^l \binom{l}{k} n(n-1) \ldots (n-k+1) (x-1)^{n-k} \frac{(-1)^{l-k} (l-k)!}{x^{1+l-k}}
 \end{eqnarray*}$$
(this last transformation thanks to Leibniz)
and since $n>l$, $f(1)=(-1)^{l+n+1}$.
In the end, your sum is equal to $(-1)^{l+n}$ if $l \geq n$, $0$ otherwise.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\sum_{m = 0}^{n}\pars{-1}^{n - m}{n \choose m}{m - 1 \choose \ell}:\
     {\large ?}.\qquad\ell \geq 0}$

\begin{align}
&\color{#66f}{\large\sum_{m = 0}^{n}\pars{-1}^{n - m}{n \choose m}
{m - 1 \choose \ell}}
\\[3mm]&=\pars{-1}^{n}\sum_{m = 0}^{n}\pars{-1}^{m}{n \choose m}
\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}{\pars{1 + z}^{m - 1} \over z^{\ell + 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\pars{-1}^{n}\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}
{1 \over z^{\ell + 1}\pars{1 + z}}
\sum_{m = 0}^{n}{n \choose m}\pars{-z - 1}^{m}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\pars{-1}^{n}\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}
{1 \over z^{\ell + 1}\pars{1 + z}}
\bracks{1 + \pars{-z - 1}}^{n}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}{1 \over z^{\ell - n + 1}\pars{1 + z}}
{\dd z \over 2\pi\ic}
=\sum_{k = 0}^{\infty}\pars{-1}^{k}\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}{1 \over z^{\ell - n - k + 1}}{\dd z \over 2\pi\ic}
\\[3mm]&=\sum_{k = 0}^{\infty}\pars{-1}^{k}\,\delta_{\ell - n,k}
=\color{#66f}{\large\left\lbrace\begin{array}{lcl}
\pars{-1}^{\ell - n} & \mbox{if} & \ell \geq n
\\[2mm]
0&&\mbox{otherwise}
\end{array}\right.}
\end{align}

