# $\sum_{k=0}^n k \binom{n}k=n\cdot2^{n-1}$

I need to prove that $\sum_{k=0}^n k \binom{n}k=n\cdot2^{n-1}$ using Combinatorial argument

I know this can be done by differentiating binomial expansion of $(1+x)^n$ and then putting $x=1$ , but how shall I do it combinatorially?

How many ways are there to colour one of $n$ balls red and paint the rest either blue or green?
First pick the red ball. Then select a subset of the remaining $n-1$ balls to paint them blue. Paint the rest green. This gives $$n\cdot 2^{n-1}.$$
Or: Pick the number $k$ of non-green balls. Pick one of these $k$ balls as red. paint the remaining $k-1$ of these balls blue. And of course the balls not picked in the first step are painted green. This gives for each $k$ a count of $k{n\choose k}$ ways, so in total $$\sum_{k=0}k{n\choose k}.$$
The left-hand side counts the number of pointed $k$-subsets of $X$ for $k=0,\cdots,n$ and the right-hand side counts the number of pointed subsets of $X$ full stop, where $|X|=n$.