How to show $ {n \choose m}{m \choose k}={n \choose k}{n-k \choose m-k}$ by counting pairs of sets $(A,B)$ in two ways and to deduce $\sum_{k=0}^n{n \choose k}{n-k \choose m-k}=2^m{n \choose m} $
I am looking for a combinatorial proof. What is a combinatorical proof: Combinatorial proof of summation of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$
So basicaly i would like to explain the sum by dissolving it into dijunctive objects with meaning.
Question: in how many ways can we form an $m$-person committee with a $k$-person subcommittee from a group of $n$ people?
Answer 1: First choose the committee in $\binom{n}{m}$ ways. Then choose the subcommittee from the committee in $\binom{m}{k}$ ways. So there are $\binom{n}{m}\binom{m}{k}$ possible such possibilities.
Answer 2: First choose the $k$ subcommittee members in $\binom{n}{k}$ ways. Then fill in the remaining $(m-k)$ committee members from the remaining pool of $(n-k)$ people. So there are $\binom{n}{k}\binom{n-k}{m-k}$ ways to do this.
Both quantities answer the same question (which only has one answer) so they must be equal. That is, $$\binom{n}{m}\binom{m}{k} = \binom{n}{k}\binom{n-k}{m-k}$$
For the second part, we use the same logic except allow the subcommittee to be any size, between $0$ and $m$ (the size of the committee). I believe you have a typo - you'll want the summation to go up to $m$, not $n$.
Then the right hand side represents first choosing the $m$ committee members, and from there picking a random subset of them to form the subcommittee. There are $2^{m}$ subsets of a set of $m$ people, so $2^{m}\binom{n}{m}$ gives the total number of possibilities.
Alternatively, you can first pick the subcommittee in $\binom{n}{k}$ ways and fill in the remaining committee members in $\binom{n-k}{m-k}$. But since $k$ is not fixed (the subcommittee can be of any size), we have to sum this over $k$ from $0$ to $m$, which gives the left hand side.
Alternatively ... using only algebra \begin{eqnarray*} \binom{n}{m} \binom{m}{k} =\frac{n!}{(n-m)!\color{red}{m!}} \frac{\color{red}{m!}}{k!(m-k)!} =\frac{n!}{k!\color{blue}{(n-k)!}} \frac{\color{blue}{(n-k)!}}{(n-m)!(m-k)!}= \binom{n}{k}\binom{n-k}{m-k} \end{eqnarray*} ... \begin{eqnarray*} \sum_{k=0}^{m} \binom{n}{k}\binom{n-k}{m-k} = \sum_{k=0}^{m}\binom{n}{m} \binom{m}{k} =\binom{n}{m} \sum_{k=0}^{m} \binom{m}{k} =\binom{n}{m} 2^m. \end{eqnarray*}
