Combinatorial proof of $\sum_{k=0}^{n} \binom{n+k-1}{k} = \binom{2n}{n}$ [duplicate]

Possible Duplicate:
Combinatorial proof for two identities

is there a combinatorial proof of equation below? (parallel summation for binomials): $$\sum_{k=0}^{n} \binom{n+k-1}{k} = \binom{2n}{n}$$
It seems like it's easy to prove combinatorically,
yet I cannot find proper proof...

The righthand side of course counts the number of $n$-element subsets of $\{1,2,\dots,2n\}$.

Now let’s classify these $n$-element subsets by their largest elements. The largest element of such a set must be one of the numbers $n,n+1,\dots,2n$, so it must be of the form $n+k$ for some $k\in\{0,1,\dots,n\}$. Suppose that you want to choose an $n$-element subset of $\{1,2,\dots,2n\}$ whose largest element is $n+k$; you can do it by choosing $n+k$ and then choosing an $(n-1)$-element subset of $\{1,2,\dots,n+k-1\}$. There are $\binom{n+k-1}{n-1}=\binom{n+k-1}k$ ways to do this, and, as previously noted, $k$ can be any integer from $0$ through $n$, so there are $\sum_{k=0}^n\binom{n+k-1}k$ ways to choose an $n$-element subset of $\{1,2,\dots,2n\}$.

Added: A slightly different way to look at it: $\binom{2n}n$ is the number of ways to seat $n$ girls and $n$ boys in a row of $2n$ seats. Suppose that the first $n-k$ seats are occupied by boys and the next seat by a girl (so this is seat $n - k + 1$). That leaves $2n-(n-k+1)=n+k-1$ seats yet to be filled. There are $k$ boys not yet seated, so there are $\binom{n+k-1}k$ ways for them to choose seats. Since $n-k$, the number of boys at the beginning of the row can be anything from $0$ through $n$, $k$ can also be anything from $0$ through $n$, and the number of possible seatings is $\sum_{k=0}^n\binom{n+k-1}k$.

First notice that $\displaystyle \binom{n+k-1}{k} = \binom{n+k-1}{n-1}$.

The right-hand side is the number of ways of picking a subset $A$ of size $n$ from a set $X$ of size $2n$. Say for argument's sake that the set we're working with is $X=\{0, 1, \cdots, 2n-1 \}$. We can split it up into cases based on the least element of $A$.

• If the least element of $A$ is $0$ then we pick the remaining $n-1$ elements from a set of size $2n-1 = n+n-1$.
• If the least element of $A$ is $1$ then we pick the remaining $n-1$ elements from a set of size $2n-2 = n+(n-1)-1$.
• $\cdots$
• If the least element of $A$ is $n$ then we pick the remaining $n-1$ elements from a set of size $n-1 = n+0-1$.

So we have the above equality.

The right side is the number of ways of choosing $n$ elements from $\{1,2,3,...,2n\}$.

The number of ways of choosing $n$ elements from that set that starting with ${1,2,...,n-k}$ and not containing ${n-k+1}$ is $\binom{n+k-1}{k}$.