Algebraic proof of $\sum_{i=0}^k{{n \choose i}{m \choose {k-i}}}= {{m+n}\choose k}$
For identities involving binomial coefficients sometimes "combinatorial proofs" and "algebraic proofs" are offered. Here we have two proofs for the Vandermonde Convolution:
Algebraic proof: The sum of the LHS of the convolution formula equals the coefficient of $x^k$ on the LHS of the polynomial $$ (1+x)^n(1+x)^m=(1+x)^{m+n}. $$ The binomial coefficient on the RHS of the convolutions formula equals the coefficient of $x^k$ on the RHS of the polynomial equation.
Combinatorial proof: Suppose that there are $n+m$ objects in a set, $n$ of them white and $m$ of them black. There are $\binom{n+m}{k}$ ways to choose $k$ elements in all. This is the RHS. The number of ways to choose $j$ white and $k-j$ black objects is the product $\binom{n}{j}\binom{m}{k-j}$, so the sum of all these objects, the LHS, must be the same total as on the RHS.
Using the binomial theorem consider $$\sum_{k=0}^{m+n}{m+n\choose k}x^k=(1+x)^{m+n}.$$ The right-hand side becomes $$(1+x)^m(1+x)^n=(\sum_{r=0}^m{m\choose r}x^r)(\sum_{s=0}^n{n\choose s}x^s).$$ The product of the two polynomials on the right-hand side can be rewritten as $$\sum_{k=0}^{m+n}\bigg{\{}\sum_{i=0}^k{n\choose i}{m\choose k-i}\bigg{\}}x^k.$$ Thus $$\sum_{i=0}^k{n\choose i}{m\choose k-i}={m+n\choose k}.$$