Trying to understand polynomials proof of Vandermonde's identity. In Spivak's calculus, page 28, problem No.4,
Asked to prove :
$$\sum_{k=0}^l {n \choose k}{m \choose l-k}={n+m \choose l}$$ by using $(1+x)^n(1+x)^m=(1+x)^{n+m}$
The answer book comes quickly from :
$$\sum_{k=0}^n {n \choose k}x^k \cdot \sum_{j=0}^m {m \choose j}x^j=\sum_{l=0}^{n+m} {n+m \choose l}x^l$$ 
to:$$\sum_{k=0}^l {n \choose k}{m \choose l-k}={n+m \choose l}$$
I was trying to expand it by definition, but doesn't work. looks like a lot of things missed in the middle, I don't understand the jump.
 A: After what you have done . For a fixed $x^l$ in right hand side,you can get$x^l$  in L.H.S by having $x^k$ from the first term and $x^{m-k}$ from the second term where $ m = 0,1,..l$. Thus now equate the coefficients to get the result.
A: Using polynomials instead of combinatorics:
$$\sum_{r=0}^{m+n} \binom{n+m}{r} x^r = 
(1+x)^{m+n} = (1+x)^{m}(1+x)^{n} =
(\sum_{i=0}^{m} \binom{m}{i}x^i)(\sum_{j=0}^{n} \binom{n}{j}x^j) = 
\sum_{r=0}^{m+n} ( \sum_{k=0}^r \binom{m}{k} \binom{n}{r-k}) x^r
$$
The coefficients must agree, and the result follows.
A: Thank you all! It's combinatoric meaning maybe obvious, but as a beginner, this algebraic step really troubled me for days. Let me try to clear my mind once again:
To see: $$\sum_{k=0}^m {m \choose k}x^k \cdot \sum_{r-k=0}^n {n \choose r-k}x^{r-k}$$
equals to : $$\sum_{r=0}^{m+n}\left(\sum_{k=0}^r {m \choose r}{n \choose r-k} \right)x^r$$
For each $?x^k \cdot ?x^{r-k}=?x^r$, their coefficient come as ${m \choose k}x^k \cdot {n \choose r-k}x^{r-k}={m \choose k}{n \choose r-k}x^r$.
