# To give a combinatoric argument for the identity $\sum_{k = 0}^{r} \binom{n - m}{k}\binom{m}{r - k} = \binom{n}{r}$ [closed]

I am finding some challenges in formulating a systematic combinatoric argument for the following equation:

$$\sum_{k=0}^r\left( \begin{array}{c} n-m \\ k \\ \end{array} \right)\left( \begin{array}{c} m \\ r-k \\ \end{array} \right) =\left( \begin{array}{c} n \\ r \\ \end{array} \right)$$ Please, assist me me.

• en.wikipedia.org/wiki/Vandermonde%27s_identity Feb 11, 2018 at 19:12
• What challenges are you finding, explicitly? This looks like your outsourcing a home work question hoping to have others do you work for you. I see description of challenges you claim you have "found", nor any attempts, nor any "dead ends" that you reached where you are stuck. Please edit your post to include your attempts, right or wrong, and to describe specifically where you are stuck. Feb 11, 2018 at 19:33

Suppose we wish to choose $r$ people from $n$, this is clearly $\binom{n}{r}$.
Now suppose there are $m$ girls and $n-m$ boys , we could choose $r-k$ girls and $r$ boys, $k$ can range over the values $k=0,1, \cdots r$, thus \begin{eqnarray*} \sum_{k=0}^{r} \binom{n-m}{k} \binom{m}{r-k} = \binom{n}{r}. \end{eqnarray*}
To choose $r$ things out of a set of $n$, break up your set of $n$ into a set of $n - m$ and a set of $m$. Choose $k$ things from the first set and $r - k$ things from the second set. You have to add up the ways to do this for each possibile value of $k$.