Method for finding sum of series $\sum_{r = 1}^{10} \binom{10}{10 - r}\binom{20}{r}$? I came across this in a problem:
$$\sum_{r = 1}^{10} \binom{10}{10 - r}\binom{20}{r}$$
Please help me solve this and give a general method to solve such sums.
 A: There isn't really a "general method", except learning how to interpret binomial coefficients, and sums and products of such.
For instance, $\binom{20}r$ may be interpreted as the number of ways to pick $r$ balls from a set of $20$ (distinguishable, numbered) balls. $\binom{10}{10-r}$ may be interpreted as picking $10-r$ balls from a different set of $10$ balls. Multiply them together, and you get the number of ways to pick $10$ balls with the condition that $r$ of those balls come from the set of $20$ (and $10-r$ of them come from the set of $10$). Sum over $r$ from $0$ to $10$, and you get the total number of ways to pick $10$ balls from the $20+10=30$ you have, without any restrictions. But this is exactly what $\binom{30}{10}$ counts. Since our sum counts the same thing as $\binom{30}{10}$, the two must be equal.
However, you're only summing from $r=1$ to $10$, so you need to subtract the $r=0$ term from the above result to get $$ \binom{30}{10}-\binom{10}{10}\binom{20}{0}=\binom{30}{10}-1$$
A: Using $$(1+x)^{20} = \binom{20}{0}+\binom{20}{1}x+\binom{20}{2}x^2+\cdots\cdots +\binom{20}{20}x^{20}.........(1)$$
Similarly $$(1+x)^{10} = \binom{10}{0}+\binom{10}{1}x^{10}+\binom{10}{2}x^{2}+\cdots \cdots \cdots+\binom{10}{10}x^{10}.....(2)$$
Now coeficients of $x^{10}$ in multiplication of $(1)$ and $(2)$
$$(1+x)^{20}\cdot (1+x)^{10} = \binom{20}{0}\cdot \binom{10}{10}+\binom{20}{1}\cdot \binom{10}{9}+\cdots \cdots +\binom{20}{10}\cdot \binom{10}{0}$$
So $$\binom{30}{10}-1 = \sum^{10}_{r=1}\binom{20}{r}\binom{10}{10-r}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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$\ds{\sum_{r = 1}^{10}{10 \choose 10 - r}{20 \choose r}:\ ?}$

It's convenient to evaluate a most general result:
\begin{align}
\sum_{r = 1}^{n}{n \choose n - r}{2n \choose r} & =
-1 + \sum_{r = 0}^{\infty}{2n \choose r}\ \overbrace{%
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{n} \over z^{n - r + 1}}
\,{\dd z \over 2\pi\ic}}^{\ds{n \choose n - r}}
\\[5mm] & =
-1 + \oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{n} \over z^{n + 1}}\
\overbrace{\sum_{r = 0}^{\infty}{2n \choose r}z^{r}
\,{\dd z \over 2\pi\ic}}^{\ds{\pars{1 + z}^{2n}}}
\\[5mm] & =
-1 +\ \underbrace{ \oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{3n} \over z^{n + 1}}\
\,{\dd z \over 2\pi\ic}}_{\ds{3n \choose n}} =\
\bbox[#ffe,10px,border:1px dotted navy]{\ds{-1 + {3n \choose n}}}
\end{align}

With $\ds{n = 10}$:

$$
\bbox[#ffe,10px,border:1px dotted navy]{\ds{\sum_{r = 1}^{10}{10 \choose 10 - r}{20 \choose r} = -1 + {30 \choose 10}} =
\color{#f00}{30,045,014}}
$$
