TITLE: Combinatorial proof for two identities
QUESTION [8 upvotes]: Does exist a combinatorial proof for the following two identities ? 

$\sum_{k = 0}^{n}  \binom{x+k}{k} = \binom{x+n+1}{n}$
$\sum_{k = 0}^{n}  k\binom{n}{k} = n2^{n-1}$


I know how to derive the identites from $(1+x)^n$ , but I am searching for a combinatorial proof ?

REPLY [4 votes]: You are choosing $n$ people from $x+n+1$ people who are lined up in a row. The right side counts the number of choices.
The left side counts the choices in another way. For any $k$ from $0$ to $n$, select all of the  first $n-k$ people (from the left), then not the next one, then choose $k$ people from the remaining $x+n+1-(n-k+1)$, that is from the remaining $x+k$.  There are $\binom{x+k}{k}$ ways to do this. The sum of the $\binom{x+k}{k}$ thus counts the choices of $n$ people from the $x+n+1$.