Proving combinatorial identity with inclusion exclusion principle

Question

After reading a bit of the inclusion exclusion principle i was trying to do some examples and got stuck on this one.

Show with the inclusion-exclusion principle that for $m,n \ge 0 $ the following identity holds

$${n \choose m } = \sum_{k=m}^n (-1)^{k-m} {k \choose m}{n \choose k } 2^{n-k} $$

Answer 1

Hint: First make the sum go backwards, that is, do the change of variable $k' = n-k$(the form becomes easier with what I am going to propose). Lets say you need a committee of at least $m$ people. The way you select the committee is by first choosing $m$ people and then the people that was not selected go to a second part of the process in which they can or cant's be part of the committee.

1. In how many ways you can form this committee using this process?
2. In how many ways you did not select anyone in the second part of the selection process?
3. Call $A_i$ be the number of committees where the $i-$th person is in the committee bu was not selected in the first part of the process.