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Prove that $$\sum\limits_{k=0}^m\binom{m}k\binom{n}{r+k}=\binom{m+n}{m+r}$$ using combinatoric arguments.

I prefer you to answer using an analogy to a combinatoric problem or else I'll have trouble understanding.

I tried:

Let $m$ represent a group of boys and $n$ represent a group of girls. We want to form a committee with the condition that there must be $r$ more girls than boys on the committee. There's two ways of doing this. (I'm having trouble with the right-hand side.)

The left-hand side is one way of doing this. So we split the problem to different cases. One such case is where there are no boys on the committee, and so, there's $\binom{m}0\binom{n}{r+0}$ ways of forming a committee. Next, we can have 1 boy on the committee. There's $\binom{m}1\binom{n}{r+1}$ ways to do this. We repeat this process until we have $\binom{m}k\binom{n}{r+k}$ ways of forming a committee. Summing up the cases will result in the right-hand side. However, as stated before, I don't really know how to explain the right-hand side. Maybe I used the wrong analogy?

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  • $\begingroup$ You could use this:math.stackexchange.com/questions/677307/… $\endgroup$ – S.C.B. Feb 17 '16 at 2:06
  • $\begingroup$ @MXYMXY I understand how to prove that question but somehow can't understand how to make sense of the right-hand side of the equation for my question. $\endgroup$ – sucksatmath Feb 17 '16 at 2:09
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Rewrite the identity using the fact that ${n \choose k} = {n \choose {n-k}}$. This yields the identity \begin{equation} \sum_{k = 0}^m {m \choose {m - k}}{n \choose {r+k}} = {{m+n}\choose{m + r}} \end{equation} Consider the right side. It is equal to the amount of ways to form a committee of size $m + r$ from $m + n$ people.

The left side can be considered in the same way. When forming a committee of size $m + n$, we can form a committee from the $m$ people and the $n$ people and then combine them. For a given value of $k$, the product ${m \choose {m - k}}{n \choose {r+k}}$ is the amount of ways to form a committee from $m + n$ people by joining two sub committees of size $m - k$ and $r + k$ yielding a committee size of $(m - k) + (r + k) = m + r$. In this way you are taking $m - k$ people from the $m$ people and $r + k$ people from the $n$ people to form a committee of size $m + r$ In order for this construction of a committee to be complete(that is it considers all possible committees) all values of $k$ must be considered, so we simply sum up the combinations for all values of $k$.

This explains both sides of the identity and concludes the combinatoric proof.

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Imagine I have a box of $m+n$ polar bears, and I want to pull $m+r$ polar bears from this box. I can do this $\binom{m+n}{m+r}$ ways. I can also split these polar bears into two boxes, one box of $m$ polar bears and another box of $n$ polar bears. I can represent the pulling of these $m+r$ polar bears by pulling $m-k$ polar bears from the box of $m$ bears and pulling $r+k$ bears from the box of $n$ bears for every $k \in [m] \cup \{0\}.$ This in essence is the same pull of bears since $m-k + r+k = m+r.$ For each $k \in [m] \cup \{0\}$, I can pull the bears from the box of $m$ bears $\binom{m}{m-k} = \binom{m}{k}$ ways and I can pull the bears from the box of $n$ bears $\binom{n}{r+k}$ ways. Hence, for each $k \in [m] \cup \{0\},$ I can pull the bears from the two boxes $\binom{m}{m-k}\binom{n}{r+k}$ ways. Thus, given how we have case $k$, $$\binom{m+n}{m+r} = \sum_{k=0}^m \binom{m}{k}\binom{n}{r+k}.$$

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