/sci/ - Science & Math

>>15216384
[eqn]\sum_{k=m}^n \binom{k}{m}\binom{n}{k} = \sum_{k=0}^{n-m} \binom{n}{m}\binom{n-m}{k} = \binom{n}{m}\sum_{k=0}^{n-m} \binom{n-m}{k} = \binom{n}{m}2^{n-m}[/eqn]

proof by pic because I don't want to type, and also I like the specific drawings in those sorts of problems.