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I saw this question here:- Combinatorial sum identity for a choose function

This looks so much like a vandermonde identity, I know we can give a counting argument for Vandermonde. However much I try I am not able to come up with a counting argument for this.

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Amrita
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  • More proofs, one of which is a counting argument, here:https://math.stackexchange.com/questions/1867972/proving-that-x-yn-1-choose-n-sum-k-0nxn-k-1-choose-n-kyk-1-ch – Mike Earnest Mar 17 '21 at 15:22

2 Answers2

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Let $S=\{-m,-m+1,\ldots,n-1,n\}$; $|S|=m+n+1$, and we want to count the subsets of $S$ of cardinality $r+s+1$. Suppose that $A$ is such a subset. Then there is a unique $k_A\in A$ such that $r$ members of $A$ are smaller than $k_A$, and $s$ members of $A$ are larger than $k_A$. Let $\mathscr{A}_k$ be the family of $(r+s+1)$-element subsets $A$ of $S$ such that $k_A=k$. There are $k-(-m)=m+k$ elements of $S$ less than $k$ and $n-k$ elements of $S$ greater than $k$, so

$$|\mathscr{A}_k|=\binom{m+k}r\binom{n-k}s\;.$$

Summing over $k$ gives us the total number of $(r+s+1)$-element subsets of $S$, which is of course $\binom{m+n+1}{r+s+1}$, so

$$\sum_{k=-m}^n\binom{m+k}r\binom{n-k}s=\binom{m+n+1}{r+s+1}\;.$$

Brian M. Scott
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  • Could you clarify this "there are k-(-m)=m+k elements of S less than k and n-k elements greater than k " ? Here k is just a element of A right? – Amrita Sep 23 '16 at 19:27
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    @Amrita: $k$ is a potential element of some $A$, so it can be any element of $S$. Since $-m$ is the smallest element of $S$, there are $k-(-m)$ elements of $S$ smaller than $k$. Similarly, since $n$ is the largest element of $S$, there are $n-k$ elements of $S$ larger than $k$. I use that information to count the sets $A\subseteq S$ such that $|A|=r+s+1$, $k\in A$, and $A$ has exactly $r$ of its elements less than $k$. – Brian M. Scott Sep 23 '16 at 19:32
  • Wow! you came up with this in 20 mins. I have been trying that question since it was posted. I would be very interested in knowing how you started and approach such a question. – Amrita Sep 23 '16 at 19:41
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    @Amrita: As you said in the question, it strongly resembles Vandermonde’s identity, so I looked for a combinatorial argument similar to the one for that identity. The terms in the summation are clearly choosing $r$ members of one set and $s$ of another, and the righthand side is clearly counting sets of size $r+s+1$, so experience tells me that there’s a good chance that the lefthand side is counting sets categorized by their $(r+1)$-st member, the one with $r$ elements below it and $s$ above. Then it was just a matter of checking that $m+k$ and $n-k$ fit that interpretation. – Brian M. Scott Sep 23 '16 at 19:46
  • Don't you think your answer would have been better suited for the original linked question? I consider this question as a duplicate. – StubbornAtom Sep 24 '16 at 06:28
  • @StubbornAtom: Since the other question has at present no combinatorial answer, and this question requires one, this one obviously is not a duplicate. It’s true that had I seen the other one first and supplied a combinatorial answer, this one would then have been unnecessary (and quite possibly even unasked, depending on the timing), but that isn’t how matters worked out. It’s a silly thing to worry about at this point anyway, since each of the questions now has a pointer to the other. – Brian M. Scott Sep 24 '16 at 07:19
  • @BrianM.Scott How do you know that this $k_A \in A$ such that $r$ members of $A$ are smaller than $k_A$ and $s$ members of $A$ are larger than $k_A$ is unique? I think I am not quite understanding what you meant there. – Sarah Sep 28 '16 at 21:23
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    @Sarah: List the $r+s+1$ elements of $A$ from smallest to largest; the $(r+1)$-st element in that list is the only member of $A$ that is larger than $r$ members of $A$ and smaller than $s$ members of $A$. – Brian M. Scott Sep 28 '16 at 21:29
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You have $m+n+2$ identical coins, and $r+s+2$ distinct boxes.
How many ways are there to place all the coins in the boxes so no box is empty?

If you place these $m+n+2$ coins in a line, then there are $m+n+1$ spaces between pairs of adjacent coins, and we can choose $r+s+1$ of these spaces to place a divider in. These dividers split the row of coins into $r+s+2$ contiguous blocks, representing the $r+s+2$ boxes, and therefore uniquely describing a placement of coins in boxes. Therefore, the number of ways to place the coins is $\binom{m+n+2}{r+s+1}$.

On the other hand, we can instead count how many ways to place the coins there are where the first $r+1$ boxes receive $m+k+1$ coins total, meaning the remaining $s+1$ boxes receive the remaining $n-k+1$ coins, and then sum over $k$. By the same argument, there are $\binom{m+k}r$ ways to place the $m+k+1$ coins in the first $r+1$ boxes, and $\binom{n-k}{s}$ ways to place the remaining $n-k+1$ coins in the other $s+1$ boxes.

Mike Earnest
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