Using a combinatorial argument I am having some difficulty with this problem:
Use a combinatorial argument to show that
$$\binom{m + n}{r} = \binom{m}{0}\binom{n}{r} + \binom{m}{1}\binom{n}{r - 1} + \dots + \binom{m}{r}\binom{n}{0}$$
My book shows how to derive an identity, but it doesn't show how to use the argument to show something. How do I go about proving this problem?
 A: There are several ways to draw r balls among m+n depending on how many you draw among the m and among the n. (r balls among the m, r-1 among the m and 1 among the n, ...)
Your result add each such case to get the total number of draws.
A: On the left you have the number of ways to pick $r$ elements from $m + n$. Of all this combinations you can split them in groups according to the number $k$ of chosen elements that fall in the first $m$ and those which fall in the last $n$.
For each value of $k$ you have $\binom{m}{k}$ of choosing those $k$ times $\binom{n}{r-k}$ of choosing the rest. You have to add up across all $k$ to reach all the combinations, and the identity holds. 
A: Let's say you had m balls in one box and n balls in the other and you mixed them up in a barrel. 
You need to select r balls from the barrel. The balls are distinguishable but the order is not important. The total ways you can do this is $\binom{m+n}{r}$
But there is another way you can calculate this. Consider the m and n ball boxes before you mixed them
You can choose 0 from the m-box and r from the n-box or 1 from the m-box and r-1 from the n-box... basically k from the m-box and r-k from the n-box for any value of k from 0 to r.
So the total number of ways you can select r balls is:
$\sum_{k=0}^{r} \binom{m}{k}\binom{n}{k-r}$
The first and second methods are the same type of selection just done differently. So you have:
$\binom{m+n}{r} = \sum_{k=0}^{r} \binom{m}{k}\binom{n}{k-r}$
A: Shuffle a deck of cards.  Draw $r$ of them.  There are $m=26$ black-suited cards and $n=26$ red-suited cards to choose from.  The first term is the number of combinations with no blacks and all reds.  The next term is the number of combinations with one black, the rest red.  Rinse and repeat until the last term has all red and no black.  The sum of all of these is the total number of combinations of cards for your $r$-card hand.
