Problem: Prove that for $0 \leq m \leq k \leq n$, $ \binom{n}{k}\binom{k}{m}=\binom{n}{m}\binom{n-m}{k-m} . $ Solution: Uh oh! Delta messed up their seating arrangements and now they need to randomly give out the $k$ upgraded (comfort plus or first class) seats to the $n$ passengers. There are $\binom{n}{k}$ ways to do this, and for each of those ways, delta needs to figure out which $m$ first class tickets they will give among the $k$ upgraded seats, and there are $k \choose m$ ways of doing this. By the multiplication principle, there are ${n \choose k}{k \choose m}$ ways to do this. Now, consider another way of giving out tickets. Delta decides to just give out the $m$ first class seats at the beginning and then give the $k-m$ comfort plus seats to the remaining $n-m$ passengers. There are ${n \choose m}{n-m \choose k - m}$ ways to do this, so the above is proven.