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

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


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


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?


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:


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



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

Consider the coefficients of $x^{20}$


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



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.