# This sigma to binom?

Can you please show me how to get from the left side to the right side?

$$\sum\limits_{k=0}^{20}\binom{50}{k}\binom{50}{20-k} = \binom{100}{20}$$

## 2 Answers

In how many ways may you pick 20 people from $100$?

$\binom{100}{20}$

If you treat the first fifty people of the hundred as special, in how many ways can you pick $k$ people from the first fifty, and $20-k$ people from the remaining fifty people?

$\binom{50}{k}\binom{50}{20-k}$

Ranging over all possible values of $k$, we see that we may count the number of ways of picking twenty people from the hundred total (with the first fifty as special) is:

$\sum\limits_{k=0}^{20}\binom{50}{k}\binom{50}{20-k}$

By combinatorial principles, if we can count the same scenario in two different ways, those answers must be equal. Therefore: $\binom{100}{20}=\sum\limits_{k=0}^{20}\binom{50}{k}\binom{50}{20-k}$

$$(x+1)^{50}(1+x)^{50}=(1+x)^{100}$$

$$(\sum_{k=0}^{50}\binom{50}k x^k)\cdot(\sum_{k=0}^{50}\binom{50}kx^{50-k})=\sum_{k=0}^{50}\binom{100}r x^r$$

Consider the coefficients of $x^{20}$

$$\binom{100}{20}$$

$$=\binom{50}0\cdot\binom{50}{20}+\binom{50}1\cdot\binom{50}{19}+\cdots+ \binom{50}{19}\cdot\binom{50}1+\binom{50}{20}\cdot\binom{50}0$$

$$=\sum_{r=0}^{20}\binom{50}r\binom{50}{20-r}$$