proving of combinational argument $\sum_{r=0}^{k}(-1)^r2^{k-r}\binom{n}{r}\binom{n-r}{k-r}= \binom{n}{k}$ How can we prove
$\displaystyle 2^k\binom{n}{0}\binom{n}{k}-2^{k-1}\binom{n}{1}\binom{n-1}{k-1}+2^{k-2}\binom{n}{2}\binom{n-2}{k-2}-.............(-1)^k\binom{n}{k}\binom{n-k}{0}=\binom{n}{k}$
$\bf{My\; Try}::$ Using Analytical Method.
We can write the given series as $\displaystyle \sum_{r=0}^{k}(-1)^r2^{k-r}\binom{n}{r}\binom{n-r}{k-r}$
$ = \displaystyle \sum_{r=0}^{k}(-1)^r2^{k-r}\frac{n!}{r!\cdot (n-r)!}\cdot \frac{(n-r)!}{(k-r)!\cdot (n-k)!}$
$= \displaystyle \sum_{r=0}^{k}(-1)^r \frac{2^k}{2^r}\cdot \frac{k!}{r!\cdot (k-r)!}\cdot \frac{n!}{k!\cdot (n-k)!} $
$= \displaystyle \binom{n}{k}\cdot 2^k\sum_{r=0}^{k}\left(-\frac{1}{2}\right)^r\cdot \binom{k}{r} = \binom{n}{k}\cdot 2^k\sum_{r=0}^{k}\left(1-\frac{1}{2}\right)^k$
$ = \displaystyle \binom{n}{k}\cdot 2^k \cdot \frac{1}{2^k} = \binom{n}{k}$
My Question is How can we solve using combinational argument
Help Required
Thanks
 A: Start with $n$ numbered white balls. How many ways are there to choose $k$ of the balls, put them in a box, throw away some subset of the balls in the box, and end up with a box containing $k$ balls?
There are $2^k\binom{n}k$ ways to pick $k$ of the balls, put them in the box, and then choose a subset of them to throw away. Now we’ll subtract the arrangements that throw away at least one ball. There are $\binom{n}1$ ways to pick the ball that’s to be thrown away, $\binom{n-1}{k-1}$ ways to pick the other $k-1$ balls to be put into the box initially, and $2^{k-1}$ ways to pick the rest of the balls that will be thrown away. This leaves
$$2^k\binom{n}k-2^{k-1}\binom{n}1\binom{n-1}{k-1}\tag{1}$$
outcomes.
Of course this overcounts, since any result that ends up with two balls being thrown away gets subtracted twice in $(1)$ and should be added back in. There are $\binom{n}2$ pairs of balls that could be the two definitely thrown away, $\binom{n-2}{k-2}$ ways to choose the other $k-2$ balls to be put into the box initially, and $2^{k-2}$ to choose from those any other balls to be thrown away. Adding all of this back in gives the approximation
$$2^k\binom{n}k-2^{k-1}\binom{n}1\binom{n-1}{k-1}+2^{k-2}\binom{n}2\binom{n-2}{k-2}\;,\tag{2}$$
and at this point, if not before, it should be clear that we’re dealing with a standard inclusion-exclusion argument, and that the number of outcomes that have all $k$ chosen balls in the box with none thrown away is
$$\sum_{r=0}^k(-1)^r2^{k-r}\binom{n}r\binom{n-r}{k-r}\;.$$
Of course this is simply the number of sets of $k$ balls, which is $\dbinom{n}k$.
