I have come across the proof for the following identiy by considering the choice function $n\choose k$ as choosing $k$ items out of $n$ items.
The identity is:
$\sum_{k=0}^n {n \choose k}{{m-n}\choose {n-k}} = {m \choose n}$
I have tried to prove this algebraically by induction but become stuck quickly, is there a simpler proof involving say binomial expansions?
If $p(x)=a_0+a_1x+\cdots+a_nx^n$ is a polynomial, let the notation $[x^k]p(x)$ refer to $a_k$, that is, the coefficient of $x^k$. (If $k$ is larger than the degree of $p(x)$, then $[x^k]p(x)=0$.)
Check that if $p(x)=a_0+a_1x+\cdots$ and $q(x)=b_0+b_1x+\cdots$ are polynomials, then the identity $$[x^m]~p(x)\cdot q(x)=\sum_{k=0}^m[x^k]p(x)\cdot[x^{m-k}]q(x)$$ holds.
Finally, determine $[x^n](1+x)^n(1+x)^{m-n}$ in two different ways.
(Note: if we instead ask for $[x^k](1+x)^n(1+x)^{m-n}$, we get a slightly more general expression.)
TL;DR What's the coefficient of $x^n$ in $(1+x)^n(1+x)^{m-n}$?
