Approach on sum of $\binom{10+r}{r}\cdot\binom{20-r}{10-r}$ The problem: Find $$\sum_{r=0}^{10} \binom{10+r}{r}\cdot\binom{20-r}{10-r}$$
I do not even know how to approach this problem. None of the standard binomials or their combinations that I consider like $(1+x)^n$ or $(1-x)^{-n}$ seem to work. Would anyone please provide a hint on how to approach this?
This is expected to be solved by only using high school techniques and considering basic polynomials of the form $(1+x)^n$ and $(1+x)^{-n}$.

Your sum is an instance of the Rothe-Hagen identity
$$\sum_{r=0}^n \frac{x}{x + rz} \binom{x + rz}{r} \frac{y}{y+(n-r)z} \binom{y+(n-r)z}{n-r} = \frac{x+y}{x+y+nz} \binom{x+y+nz}{n} \tag{1}$$ for the choice $$x = y = 11, \quad n = 10, \quad z = 1.$$  In particular, this yields
$$\begin{align}
\sum_{r=0}^{10} \binom{10+r}{r} \binom{20-r}{10-r} 
&= \sum_{r=0}^{10} \frac{11}{11+r} \binom{11+r}{r} \frac{11}{21-r} \binom{21-r}{10-r} \\
&= \frac{22}{32} \binom{32}{10} \\
&= \binom{31}{10}.
\end{align}$$
