This sigma to binom? Can you please show me how to get from the left side to the right side? 
$$\sum\limits_{k=0}^{20}\binom{50}{k}\binom{50}{20-k} = \binom{100}{20}$$ 
 A: 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: $\binom{100}{20}=\sum\limits_{k=0}^{20}\binom{50}{k}\binom{50}{20-k}$
A: $$(x+1)^{50}(1+x)^{50}=(1+x)^{100}$$
$$(\sum_{k=0}^{50}\binom{50}k x^k)\cdot(\sum_{k=0}^{50}\binom{50}kx^{50-k})=\sum_{k=0}^{50}\binom{100}r x^r$$
Consider the coefficients of $x^{20}$
$$\binom{100}{20}$$
$$=\binom{50}0\cdot\binom{50}{20}+\binom{50}1\cdot\binom{50}{19}+\cdots+
\binom{50}{19}\cdot\binom{50}1+\binom{50}{20}\cdot\binom{50}0$$
$$=\sum_{r=0}^{20}\binom{50}r\binom{50}{20-r}$$
