Is this combinatorics summation equality true? Recently I came upon the following equality, for natural numbers $n,k$ such that $n\ge k\ge0$:
$$\binom nk=\sum_{m=0}^{\min(k,n-k)}\binom km\binom{n-k}m$$
First of all, is this equality even true? It seems to be true from the numbers I have checked, but I am not 100% sure. Also, not sure if this is some well known theorem or identity so sorry in advance if it is.
Regardless, I don't really have an idea of how to approach a problem like this(the $\min(k,n-k)$ part complicates it for me especially). Should I try to create a hypothetical scenario to fit this equation? Or should I use some combinatoric identities to arrive at a conclusion?
Thanks for the help.
 A: Let $k$ be a nonnegative integer. The famous Vandermonde identity
$$\binom{x+y}k=\sum_{m=0}^k\binom xm\binom y{k-m}$$
has an obvious combinatorial interpretation when $x$ and $y$ are nonnegative integers, and holds for all values of $x$ and $y$ since both sides are polynomials of degree $k$ in $x$ and $y$.
Setting $x=n-k$ and $y=k$ we get
$$\binom nk=\sum_{m=0}^k\binom{n-k}m\binom k{k-m}=\sum_{m=0}^k\binom km\binom{n-k}m.$$
This identity holds for all real or complex values of $n$. Note that, if $n$ is an integer and $n\ge k$, then $n-k$ is a nonnegative integer and the binomial coefficient $\binom{n-k}m$ vanishes for $m\gt n-k$. In this case the upper limit of summation can be changed to $\min(k,n-k)$.
A: Here is famous VanderMonde equality, $$\binom{N_1+N_2}{k}=\sum_{m=0}^{N_1}\binom{N_1}{m}\binom{N_2}{k-m}, k<=N_1+N_2$$,
let us do a replacement, $n=N_1+N_2$,and if $k=N_2$, we get $$\binom{n}{k}=\sum_{m=0}^{n-k}\binom{n-k}{m}\binom{k}{k-m}=\sum_{m=0}^{n-k}\binom{n-k}{m}\binom{k}{m}$$
obviously in above equality, $m<=k$, so if $k<n-k$, the sum at most reach to $\min(k,n-k)$, so can get the equality:
$$\binom{n}{k}=\sum_{m=0}^{\min(k,n-k)}\binom{n-k}{m}\binom{k}{m}$$
