<s>[INST] Proof strategy to show $\large \sum_{0 \ \le \ x \ \le \ a} \normalsize \binom{a}{x} \binom{b}{n+x} = \binom{a+b}{a+n} $ The following two combinatorial identities are taken from a textbook. 
$$\begin{align}
&\large \sum_{0\ \le \ x \ \le \ n} \normalsize \binom{a}{x} \binom{b}{n-x} = \binom{a+b}{n} \tag{10} \\
&\large \sum_{0 \ \le \ x \ \le \ a} \normalsize \binom{a}{x} \binom{b}{n+x} = \binom{a+b}{a+n}\tag{11} \\
\end{align}$$ 
What proof strategy would you suggest to prove identity $(11)$, algebraically?  We may assume Vandermonde's Identity $(10)$; it may be useful in a proof of identity $(11)$.  
There appear to be many similarities between the two identities. How might we make use of that?
 [/INST] Identity (11), just like (10), is obvious on combinatorial grounds, but you asked for an algebraic proof. Let $y=a-x$.$$\binom{a+b}{a+n}=\sum_{x=0}^\infty\binom ax\binom b{a+n-x}=\sum_{x=0}^\infty\binom a{a-x}\binom b{a-x+n}=\sum_{y=0}^\infty\binom ay\binom b{y+n}=\sum_{x=0}^\infty\binom ax\binom b{x+n}$$
</s>