Proof for Vandermonde's identity: ${{m+n} \choose r} = \sum_{k=0}^r {m \choose k}{n\choose {r-k}}$ There have been a number of proofs for this, such as: 
How to prove Vandermonde's Identity: $\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}$?, and
Vandermonde's Identity: How to find a closed formula for the given summation.
However, they involve a lot of hand-waving or "consider a K-by-K matrix.. or "Suppose a committee consists of m men and n women".
I'm looking for a good solid step-by-step proof:
The best one I've found comes from this's Algebraic Proof:

where in line 1, the binomial theorem is applied. 
From lines 1 to 2, it's just a factoring of exponents. 
From line 2 to 3, it's an application of the binomial theorem on each factor term.
From line 3 to 4 however, I'm looking for some missing (assumed obvious) steps. They'll probably be a set of change of variables such as $\textrm{let } j=r-i$ then a new equation, then maybe another change of variables.
Can someone please provide step-by-step equations from line 3 to line 4?
 A: Simple rearranging gives
\begin{equation}\begin{aligned}
\sum_{i=0}^m\begin{pmatrix}m\\i\end{pmatrix}x^i\sum_{j=0}^n\begin{pmatrix}n\\j\end{pmatrix}x^j&=\sum_{i=0}^m\sum_{j=0}^n\begin{pmatrix}m\\i\end{pmatrix}\begin{pmatrix}n\\j\end{pmatrix}x^{i+j}.
\end{aligned}\end{equation}
The trick  is to then swap the sum over $i$ and $j$ for a sum over $r=i+j$ and $i$.
\begin{equation}\begin{aligned}
\sum_{i=0}^m\begin{pmatrix}m\\i\end{pmatrix}x^i\sum_{j=0}^n\begin{pmatrix}n\\j\end{pmatrix}x^j&=\sum_{r=0}^{m+n}\,\sum_{i=0}^{\min(m, r)}\begin{pmatrix}m\\i\end{pmatrix}\begin{pmatrix}n\\r-i\end{pmatrix}x^{r}.
\end{aligned}\end{equation}
Finally, we use that $\begin{pmatrix}a\\b\end{pmatrix} = 0$ for $b>a$ to extend the sum over $i$ from $\min(m, r)$ to $r$.
\begin{equation}\begin{aligned}
\sum_{i=0}^m\begin{pmatrix}m\\i\end{pmatrix}x^i\sum_{j=0}^n\begin{pmatrix}n\\j\end{pmatrix}x^j&=\sum_{r=0}^{m+n}\,\sum_{i=0}^r\begin{pmatrix}m\\i\end{pmatrix}\begin{pmatrix}n\\r-i\end{pmatrix}x^{r}.
\end{aligned}\end{equation}
A: 
A step-by-step derivation:  We obtain
  \begin{align*}
\color{blue}{\left(\sum_{i=0}^m\right.}&\color{blue}{\left.\binom{m}{i}x^i\right)\left(\sum_{j=0}^n\binom{n}{j}x^j\right)}\\
&=\sum_{i=0}^m\sum_{j=0}^n\binom{m}{i}\binom{n}{j}x^{i+j}\tag{1}\\
&=\sum_{r=0}^{m+n}\sum_{{i+j=r}\atop{i,j\geq 0}}\binom{m}{i}\binom{n}{j}x^r\tag{2}\\
&\,\,\color{blue}{=\sum_{r=0}^{m+n}\left(\sum_{i=0}^r\binom{m}{i}\binom{n}{r-i}\right)x^r}\tag{3}
\end{align*}

Comment:


*

*In (1) we multiply out and collect the $x$-terms.

*In (2) we rearrange the terms by increasing powers of $x$.

*In (3) we substitute $j=r-i$ eliminating $j$ this way.
A: Both linked questions have an answer expressing that the formula in your title is for all practical purposes obvious. At any rate, a proof does not require calculations unpacking the expressions appearing in this formula. Just select  $r$ people from a set of $m$ men and $n$ women.
