$\sum_{i=0}^m \binom{m-i}{j}\binom{n+i}{k} =\binom{m + n + 1}{j+k+1}$ Combinatorial proof Is there a simple combinatorial proof for the following identity?
$$\sum_{0\leq i \leq m} \binom{m-i}{j}\binom{n+i}{k} =\binom{m + n + 1}{j+k+1}$$
where $m,j \geq 0$, $k \geq n \geq 0$.
 A: Count $(j+k+1)$ element subsets of an ordered set with ($m+n+1$) elements, organizing the subsets by the value of the $(j+1)$-th largest element in the subset.  The product of two binomial coefficients on the left side of the equation is the number of ways to choose the rest of the subset when the $(j+1)$-th largest elements in the subset is the $(m-i)$-th largest element in the whole set.
I think this formula is called the dual Vandermonde or negative Vandermonde identity.
Edit.  The inequality between $n$ and $k$ in the question has the wrong sign, it should be $n \leq k$.
A: (Essentially the same as the above)
Counting the number of lattice paths of length l+j+k+1 from $(0,0)$ to $(j+k+1,l)$ with steps $(0,1)$ and $(1,0)$ in two ways, we can derive this identity. Each path  includes $(j,s)$ to $(j+1,s)$ step $(s=0,1\ldots l)$.In this way, we can divide paths into l+1 groups which are mutually exclusive. Hence,
$\sum_{0\leq s \leq l} \binom{s+j}{j}\binom{l-s+k}{k} =\binom{l+j+k+1}{l}$
Setting $l=m+n-(j+k)$, $s=(m-j-i)$, we have the identity in question. 
