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

Question

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.

Answer 1

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*}

Answer 2

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 $k$.