Approach on sum of $\binom{10+r}{r}\cdot\binom{20-r}{10-r}$

Question

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 $(1+x)^{-n}$.

Answer 1

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 $$\binom{31}{21} = \binom{31}{10} = \boxed{44352165}.$$

Answer 2

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 $$\begin{align} \sum_{r=0}^{10} \binom{10+r}{r} \binom{20-r}{10-r} &= \sum_{r=0}^{10} \frac{11}{11+r} \binom{11+r}{r} \frac{11}{21-r} \binom{21-r}{10-r} \\ &= \frac{22}{32} \binom{32}{10} \\ &= \binom{31}{10}. \end{align}$$

Answer 3

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 $(1+x)^s(1+x)^t=(1+x)^{s+t}$.