Use of generating function in combinatorial proof In Combinatorial proof of summation of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$, the product 
$\left(\sum_{a=0}^n\binom{n}{a}x^a\right)\left(\sum_{b=0}^n\binom{n}{b}x^{n-b}\right)=\sum_{c=0}^{2n}\left(\sum_{a+n-b=c}\binom{n}{a}\binom{n}{b}\right)x^c$
This approach uses a "generating function" and collects all the coefficients of $x^{c}$ for fixed $c$ into $\sum_{a+n-b=c}\binom{n}{a}\binom{n}{b}$,  to get $\sum_{c=0}^{2n}\left(\sum_{a+n-b=c}\binom{n}{a}\binom{n}{b}\right)x^c$
I know that $\left(\sum_{a=0}^n\binom{n}{a}x^a\right)\left(\sum_{b=0}^n\binom{n}{b}x^{n-b}\right)= \sum_{a=0}^{n}\sum_{b=0}^{n}\binom{n}{a}\binom{n}{b}x^{a+n-b}$, but it remains unclear how one can obtain  $\sum_{c=0}^{2n}\left(\sum_{a+n-b=c}\binom{n}{a}\binom{n}{b}\right)x^c$ from the initial product. 

Here we also have an application of the Cauchyproduct of series.

\begin{align*}
\left(\sum_{a=0}^n\binom{n}{a}x^a\right)\left(\sum_{b=0}^n\binom{n}{b}x^{n-b}\right)
=\sum_{c=0}^{\color{blue}{2n}}\left(\sum_{\color{blue}{{a+(n-b)=c}}\atop{a\geq 0,n-b\geq 0}}\binom{n}{a}\binom{n}{b}\right)x^c\tag{1}
\end{align*}



*

*We have the (blue marked) upper limit $2n$ which comes from $x^ax^{n-b}=x^{a+n-b}$ if $a=(n-b)=n$, i.e. $a=n$ and $b=0$.

*We have $x^c=x^ax^{n-b}$ with $0\leq c\leq 2n$. In order to get the coefficient $x^c=x^{a+(n-b)}$ we have the (blue marked) condition $a+(n-b)=c$ as lower limit in the inner sum.
