Binomial coefficient equality

Question

QUESTION: \begin{equation*} \dbinom{n+m+1}{n}=\sum_{k=0}^n \dbinom{r+k}{k} \dbinom{m+n-r-k}{n-k} \end{equation*} where $n, m, r \geq 0$

I tried proving using binomial coefficient formula $\big[\dbinom{n}{k}=\frac{n!}{k!(n-k)!}\big]$, but dont think its possible with summation, however if it is would it work? Maybe lattice paths would work better, but my understanding of lattice squares is minimal.

So i attempted the proof using binomial theorems and exponent combination laws and i get stuck

\begin{equation*} \sum_{k=0}^n \dbinom{r+k}{k} \dbinom{m+n-r-k}{n-k}=\dbinom{n+m+1}{n} \end{equation*} working with the right side . \begin{equation*}\dbinom{n+m+1}{n}= \sum_n\dbinom{n+m+1}{n}x^n=(1+x)^{n+m+1}\end{equation*} \begin{equation*}=(1+x)^n(1+x)^m(1+x)\end{equation*} or would the next step be \begin{equation*}(1+x)^n(1+x)^{m+1}\end{equation*}

I think the LHS should be $\dbinom{n+m}n$, in which case this is just a simple variation of Vandermonde's identity. — Prasun Biswas, Nov 03 '17 at 02:00
@PrasunBiswas no, the LHS is correct: this a "double convolution" (upper and lower terms), in which case the upper term gets a +1. — G Cab, Nov 03 '17 at 09:24
@MahlissaJayde: The right-hand side $(1+x)^{n+m+1}$ of your calculation is not valid, since $n$ is the index of the sum and so $n$ can't occur on the right-hand side. — Markus Scheuer, Nov 03 '17 at 22:32

Markus Scheuer · Accepted Answer · 2017-11-03T22:38:16.640

2

We obtain \begin{align*} \color{blue}{\sum_{k=0}^n}&\color{blue}{\binom{r+k}{k}\binom{m+n-r-k}{n-k}}\\ &=\sum_{k=0}^n\binom{-r-1}{k}(-1)^k\binom{-m+r-1}{n-k}(-1)^{n-k}\tag{1}\\ &=(-1)^n\binom{-m-2}{n}\tag{2}\\ &\color{blue}{=\binom{n+m+1}{n}}\tag{3} \end{align*}

Comment:

In (1) we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ twice.

This way we get for example $\binom{r+k}{k}=\binom{-(-r-k)}{k}=\binom{(-r-k)+k-1}{k}(-1)^k=\binom{-r-1}{k}(-1)^k$ with $-r-k=p$ and $k=q$.
In (2) we apply the Chu-Vandermonde identity.
In (3) we apply again the binomial identity as in (1).

edited Nov 03 '17 at 22:38

answered Nov 03 '17 at 09:19

Markus Scheuer

108,315

perfect (as always!) – G Cab Nov 03 '17 at 09:27
@GCab: You're too friendly, thanks. But sometimes there are nice miniatures, which I don't see straightaway. This post is a nice example where SangchulLee provided a very nice answer to my hint. :-) – Markus Scheuer Nov 03 '17 at 09:32
Markus, yes, I saw that elegant shortcut provided by Sangchul Lee: that is the nice aspect of this site, learning from each other. And in this regard I would like to know your approach to this other interesting post if I may permit to abuse of your "intellectual curiosity" . – G Cab Nov 03 '17 at 15:22
thank you, ive always liked algebra until it became abstract – Mahlissa LECKY Nov 03 '17 at 17:52
@MahlissaJayde: You're welcome! Good to see the answer is helpful! :-) – Markus Scheuer Nov 03 '17 at 21:34
i never learned chu vandermonde, is this the only way ? i clicked the link and i saw upside L's and something about Gauss, where there was no further information – Mahlissa LECKY Nov 03 '17 at 22:07
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@MahlissaJayde: There's always more than one way to do it. Not so clear what you want to know. Some more information to Chu Vandermonde or alternative proofs of your problem? – Markus Scheuer Nov 03 '17 at 22:14
well i just edited my question with my attempt of proving, but then i end up at a dead end – Mahlissa LECKY Nov 03 '17 at 22:24
unless i'm over thinking and the chu vandermonde identity is just a method that lets you apply operations on binomial identities. Correct me if im wrong please, multiplied $(-1)^k(-1)^{n-k}$ and somehow got the difference of $\dbinom{-r-1}{k}$ and $\dbinom{-m+r-1}{n-k}$ – Mahlissa LECKY Nov 03 '17 at 22:29
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@MahlissaJayde: I've extended my comment somewhat which might be useful. – Markus Scheuer Nov 03 '17 at 22:38
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i understood that step, thank you its just the elimination of k in step 2 – Mahlissa LECKY Nov 03 '17 at 22:41

Binomial coefficient equality

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