Binomial ID $\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=\binom{n+p-m}{n}$ Could I get some help with proving this identity?
$$\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=\binom{n+p-m}{n}.$$
It has been checked in Matlab for various small $n,m$ and $p$. I have a proof for $m=0$ that involves Pascal's rule to split it into two sums that mostly cancel, but this does not work for all $m$. I have looked at induction on various letters (and looked at simultaneous induction on multiple letters) without any success. 
I would appreciate any help, algebraic or combinatoric.
 A: It's a generating-function exercise.  We have
$$
\sum_{k=m}^\infty (-1)^{k+m} {k \choose m} x^{k-m} = (1+x)^{-(m+1)},
$$
and (with $j=p-k$)
$$
\sum_{j=0}^\infty {n+p+1 \choose n+(p-j)+1} x^j
= \sum_{j=0}^\infty {n+p+1 \choose j} x^j = (1+x)^{n+p+1}.
$$
The desired sum is the $x^{p-m}$ coefficient of the product 
$(1+x)^{n+p-m}$ of these two power series, which is 
${n+p-m \choose p-m} = {n+p-m \choose n}$, QED.
A: I wish to explain a modern (high tech) method called the Wilf-Zeilberger (WZ) technique which might help you (and anyone interested) with the present question and many others you encounter in the future. This will save you time from hunting the literature and comparing notes with the milliard hypergeometric formulas.
Zeilberger developed an accompanying algorithm package which is by now part of the symbolic softwares, Maple and Mathematica. In case you have access to Maple, you may download and run a freely available online copy of it from here.
Start with the discrete function
$$F(p,k)=(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}\binom{n+p-m}{n}^{-1}.$$
The above-mentioned algorithm furnished the companion function
$$G(p,k)=-\frac{F(p,k)(k-m)(n+k+1)}{(p+1-k)(n+p+1-m)}.$$
Check (preferable using a symbolic software) that
$$F(p+1,k)-F(p,k)=G(p,k+1)-G(p,k).\tag1$$
Convention: $\binom{a}b=0$ if $b>a$ or $b<0$. Now, sum both sides of (1) over all integers $k$, i.e. $\sum_{-\infty}^{\infty}$. Notice that the RHS of (1) vanishes because $\sum_{k\in\mathbb{Z}}G(p,k+1)=\sum_{k\in\mathbb{Z}}G(p,k)$. If we denote $f(p):=\sum_{k=m}^pF(p,k)=\sum_{\mathbb{Z}}F(p,k)$ then the LHS of (1) implies
$$f(p+1)=f(p).$$
But, if $p=0$ then $f(0)=1$ and hence $f(p)=1$ identically. That means (after rewriting)
$$\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=\binom{n+p-m}{n}$$
as desired.
A: Such identities are often reduced to the Chu--Vandermonde's identity $\sum_{i+j=\ell} \binom{x}i\binom{y}j=\binom{x+y}\ell$ by using reflection formulae $\binom{x}k=\binom{x}{x-k}$, $\binom{x}k=(-1)^k\binom{k-x-1}k$.
In your case you may write
$$
\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=
\sum_{k=m}^p \binom{-m-1}{k-m}\binom{n+p+1}{p-k}=
\binom{n+p-m}{p-m}
$$
as you need, so it is Chu--Vandermonde for $x=-m-1$, $y=n+p+1$, $\ell=p-m$.
A: Mathematica says: 
$$
(-1)^{2 m} \binom{n+p+1}{m+n+1} \, _2F_1(m+1,m-p;m+n+2;1)
$$
