# alternative proof to induction, binomial sum identity

$\displaystyle \sum_{q=0}^{k} \begin{pmatrix}n-1+q\\ n-1 \end{pmatrix} = \begin{pmatrix}n+k\\n \end{pmatrix}$

induction

$\begin{pmatrix}n \\ k \end{pmatrix} := \frac{n!}{(n-k)!k!}$

Beginning of induction : $k=0 \rightarrow \begin{pmatrix} n-1 \\ n-1 \end{pmatrix} = \begin{pmatrix} n \\ n \end{pmatrix} = 1$

Induction step : $k\rightarrow k+1: \sum_{q=0}^{k+1} \begin{pmatrix} n-1+q\\n-1 \end{pmatrix} = (\sum_{q=0}^{k} \begin{pmatrix} n-1+q\\n-1 \end{pmatrix}) + \pmatrix{n-1+k+1\\n-1}$

$\displaystyle = \begin{pmatrix}n+k \\ n \end{pmatrix} + \begin{pmatrix} n+k \\ n-1 \end{pmatrix} = \frac{(n+k)!}{k!(n)!}+ \frac{(n+k)!}{(k+1)!(n-1)!} =$

$\displaystyle =\frac{(n+k)!(k!n!+(k+1)k!(n-1)!}{k!n!(k+1)!(n-1)!} = \frac{(n+k)!(n+k+1)}{n!(k+1)!} = \begin{pmatrix}n+k+1 \\ n \end{pmatrix}$

Do you know any other way to show this? Please do tell.

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Your question is phrased in an inappropriate manner. No one here has to offer anything to you if they don't want to - ask politely. –  Zev Chonoles Nov 22 '11 at 18:15
sorry ………………... –  VVV Nov 22 '11 at 18:17
This is much better, thank you for editing. I have upvoted your question. –  Zev Chonoles Nov 22 '11 at 18:23
It is likely that the problem lies with the translation into English of a very polite way of asking a question. –  André Nicolas Nov 22 '11 at 18:24

$\binom{n+k}{n}$ is the number of ways of choosing $n$ objects from $n+k$ possibilities.
In order to select $n$ objects out of $n+k$ possibilities, we can also proceed as follows: number the objects from $1$ throught $n+k$. First, pick the object with the largest number that you will select; it must be numbered somewhere between $n$ and $n+k$ (otherwise, you cannot pick $n$ objects total); say it is numbered $n+q$, with $q\in\{0,\ldots,k\}$. Then you still have to pick $n-1$ objects from the first $n+q-1$ objects; this can be done in precisely $\binom{n+q-1}{n-1}$ ways.
Since you can do this for each value of $q\in\{0,\ldots,k\}$, adding up these possibilities will result in a selection of $n$ objects from $n+k$ possibilities; that is, $$\sum_{q=0}^{k}\binom{n+q-1}{n-1} = \binom{n+k}{n}.$$