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.

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 $$\binom{30}{10}-1 = \sum^{10}_{r=1}\binom{20}{r}\binom{10}{10-r}$$
