Proof Concerning Sum of Binomial Coefficients

Question

Could anybody provide a proof of the following identity identity:

$$ \sum_{n=0}^{N-1}\binom{N-1+n}{n}=\binom{2N-1}{N}$$

possibly using Symmetry property and Pascal's rule (or another easier way):

$$\binom{a}{b}=\binom{a-1}{b-1}+\binom{a-1}{b}$$

Answer 1

$$\begin{eqnarray} \sum_{k=0}^N {N+k \choose N} &=& \sum_{k=0}^N {N-1+k \choose N-1} + \sum_{k=-1}^{N-1} {N+k \choose N} \\ &=& \sum_{k=0}^{N-1} {N-1+k \choose N-1} + \sum_{k=0}^N {N+k \choose N} + {2N-1 \choose N-1} - {2N \choose N} \end{eqnarray}$$

So

$$\begin{eqnarray} 0 &=& \sum_{k=0}^{N-1} {N-1+k \choose N-1} + {2N-1 \choose N-1} - {2N \choose N} \\ &=& \sum_{k=0}^{N-1} {N-1+k \choose N-1} - {2N - 1 \choose N} \end{eqnarray}$$

Answer 2

Note that $${2N-1 \choose N}$$ is the number of ways to choose $N$ distinct numbers from the set $\{1,2,\ldots, 2N-1\}$. Group the possibilities by the largest of these $N$ numbers. This largest number is at least $N$. For $k \in \{0, \ldots, N-1\}$ there are $${N - 1 + k \choose N-1}$$ cases where the largest number is $N+k$. So

$$ \sum_{k=0}^{N-1} {N-1+k \choose k} = \sum_{k=0}^{N-1} {N-1+k \choose N-1} = {2N-1 \choose N}.$$