Prove that $$ \binom{n}{m}\binom{m}{m} + \binom{n}{m+1}\binom{m+1}{m} + \binom{n}{m+2}\binom{m+2}{m} + \cdots + \binom{n}{n}\binom{n}{m} = \binom{n}{m} 2^{n-m}$$using a committee forming argument.
I think we can rewrite this as $\displaystyle\sum_{k=m}^{k=n} \binom{n}{k}\binom{k}{m}+\binom{n}{k+r}\binom{k+r}{m}=\binom{n}{m}\cdot2^{n-m}.$
Here is what I have so far: We are given a group of $n$ people, in which we need to make a committee of a minimum of $m$ people. From the committee, we are choosing a subset of $m$ people. How would I continue? Thanks in advance.
Let's form a committee of at least $m$ people and a subcomittee thereof comprising exactly $m$ people. Our number of options is the original sum of products of pairs of binomial coefficients. Equivalently, choose the subcommittee first in one of $\binom{n}{m}$ ways, and decide who if anyone to add to the ancillary part of the committee in one of $2^{n-m}$ ways.