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 $(1+x)^{-n}$.
 A: First off, use the identity
$$\binom{n}{k} = \binom{n}{n - k}$$
to simplify the sum to
$$\sum_{r = 0}^{10} \binom{10 + r}{10} \binom{20 - r}{10}.$$
Now, using the convention that $\binom nk = 0$ if $k < 0$ or $k > n$, this can be simplified even further into
$$\sum_{r = -10}^{20}\binom{10 + r}{10} \binom{20 - r}{10} = \sum_{s = 0}^{30} \binom{s}{10}\binom{30 - s}{10},$$
where we've made the substitution $s = r + 10$. This sum count up the number of ways we can split a line of $30$ balls into a left and right group and choose $10$ balls from each group. Another way to do this would be to choose $21$ balls out of a line of $31$ balls, and split the groups according to the $11$th ball chosen (removing it from the line). This means that the answer is $$\binom{31}{21} = \binom{31}{10} = \boxed{44352165}.$$
A: 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
$$\begin{align}
\sum_{r=0}^{10} \binom{10+r}{r} \binom{20-r}{10-r} 
&= \sum_{r=0}^{10} \frac{11}{11+r} \binom{11+r}{r} \frac{11}{21-r} \binom{21-r}{10-r} \\
&= \frac{22}{32} \binom{32}{10} \\
&= \binom{31}{10}.
\end{align}$$
A: Here is an approach using two common binomial identities which might also be of interest.

We obtain
\begin{align*}
\color{blue}{\sum_{r=0}^{10}}\color{blue}{\binom{10+r}{r}\binom{20-r}{10-r}}
&=\sum_{r=0}^{10}\binom{-11}{r}\binom{-11}{10-r}(-1)^r(-1)^{10-r}\tag{1}\\
&=\binom{-22}{10}\tag{2}\\
&\,\,\color{blue}{=\binom{31}{10}=44\,352\,165}
\end{align*}

Comment:

*

*In (1) we use the binomial identity $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$.


*In (2) we apply the Chu-Vandermonde identity
$\sum_{k=0}^n\binom{s}{k}\binom{t}{n-k}=\binom{s+t}{n}$ which corresponds to $(1+x)^s(1+x)^t=(1+x)^{s+t}$.
