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Does someone know, if the subsequent formula holds for $m \ge n \ge i \ge 1$ and if yes, can give a reference. $$\sum_{k=i}^{m-n+i}\binom{k}{i}\binom{m-k}{n-i} = \binom{m+1}{n+1}$$ Thank you very much!

Martin Sleziak
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Ralph
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2 Answers2

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What you want is equation (5.26) on page 169 of Concrete Mathematics (2nd edition) by Ronald Graham, Donald Knuth, and Oren Patashnik.

Corrected: For integers $m,n\geq0$ and integers $\ell,q$ with $\ell+q\geq 0$ we have $$\sum_{-q\leq k\leq \ell}{q+k\choose n}{\ell-k\choose m}={\ell+q+1\choose m+n+1}.$$

Let's now substitute your variables $i\geq 0$ and $n-i\geq 0$ in the bottom to obtain $$\sum_{-q\leq k\leq \ell}{q+k\choose i}{\ell-k\choose n-i}={\ell+q+1\choose n+1}.$$

In fact, since you have assumed $i>0$, we get even more $$\sum_{i-q\leq k\leq \ell}{q+k\choose i}{\ell-k\choose n-i}={\ell+q+1\choose n+1}.$$ That's because ${q+k\choose i}=0$ when $q+k<i$.

Redefining the $k$ variable gives $$\sum_{i\leq k\leq \ell+q}{k\choose i}{\ell+q-k\choose n-i}={\ell+q+1\choose n+1}.$$

Letting $m=\ell+q$ gives $$\sum_{i\leq k\leq m}{k\choose i}{m-k\choose n-i}={m+1\choose n+1}.$$

Is this the same as your sum? Yes!

  1. If $n=i$, then this is obvious.

  2. When $n>i$, then ${m-k\choose n-i}=0$ for $k>m-n+i$ anyway.

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    Byron, there are a couple of typos in the book for that sum. The summation should be for $-q \leq k \leq \ell$, and it's valid for integers $m,n \geq 0$ and integer $\ell + q \geq 0$. (The corrections are listed on Knuth's web site.) – Mike Spivey Jan 12 '12 at 16:30
  • @Mike Yikes! I should have checked that. Thanks. –  Jan 12 '12 at 16:38
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    Knuth's books tend to have few errors; I was surprised when I discovered this one. – Mike Spivey Jan 12 '12 at 16:42
  • Thanks to you both, Byron and Mike. But I think both sums are not quite the same: In Knuth's formula there are $l+q+1$ summands (equals the "nominator" in the right-hand-side binomial), while in mine there are $m-n+1$ sumands corresponding to the difference of the "nominator" and "denominator" of the r.h.s-binomial. Accordingly, I was able to transform Knuth's formula into $$\sum_{k=0}^m\binom{k}{i}\binom{m-k}{n-i}=\binom{m+1}{n+1}$$ – Ralph Jan 12 '12 at 19:32
  • Anyway, my formula can be proved by a quite simple induction. Tanks again. – Ralph Jan 12 '12 at 19:49
  • @Ralph I'm glad you found your answer. Anyway, I have added the change of variables argument, just for fun. –  Jan 12 '12 at 20:22
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    @MikeSpivey, did you claim your $2.18? – Bruno Joyal Jan 12 '12 at 22:17
  • @Byron: Yeah, sure. Thanks for showing. I should also have seen that in the formula in my comment above one can assume $k \ge i$ (otherwise first binom is zero) and $m-k \ge n-i$ (otherwise 2nd binom zero). – Ralph Jan 12 '12 at 23:51
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    @Bruno: I wasn't the first to discover it, so no. :) I do have a check from Knuth for finding an error in The Art of Computer Programming, though. – Mike Spivey Jan 12 '12 at 23:55
  • @Ralph Careful with that second one, you can only assume that $m-k\geq n-i$ provided that $n-i>0$. –  Jan 12 '12 at 23:57
  • @MikeSpivey Ha ha! I also have a check from Knuth. It's framed and on my office wall. –  Jan 12 '12 at 23:59
  • Byron, mine is framed as well. The check itself is worth much more than the money. :) – Mike Spivey Jan 13 '12 at 00:52
  • Mike and Byron - I am jealous. It's ironic that I wrote $2.18 above, and no-one corrected me. :) – Bruno Joyal Jan 13 '12 at 05:11
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = i}^{m - n + i}{k \choose i}{m - k \choose n - i} & = \sum_{k = 0}^{\infty}{k + i \choose i}{m - k - i \choose n - i} = \sum_{k = 0}^{\infty}{k + i \choose k}{m - k - i \choose m - k - n} \\[5mm] & = \sum_{k = 0}^{\infty}\bracks{{-i - 1 \choose k}\pars{-1}^{k}} \bracks{{i - n - 1\choose m - k - n}\pars{-1}^{m - k - n}} \\[5mm] & = \pars{-1}^{m - n}\sum_{k = 0}^{\infty}{-i - 1 \choose k} \bracks{z^{m - k - n}}\pars{1 + z}^{i - n - 1} \\[5mm] & = \pars{-1}^{m - n}\bracks{z^{m - n}}\pars{1 + z}^{i - n - 1}\sum_{k = 0}^{\infty}{-i - 1 \choose k}z^{k} \\[5mm] & = \pars{-1}^{m - n}\bracks{z^{m - n}}\pars{1 + z}^{i - n - 1}\, \pars{1 + z}^{-i - 1} = \pars{-1}^{m - n}\bracks{z^{m - n}}\pars{1 + z}^{-n - 2} \\[5mm] & = \pars{-1}^{m - n}{-n - 2 \choose m - n} = \pars{-1}^{m - n}\bracks{{m + 1 \choose m - n}\pars{-1}^{m - n}} = \bbx{m + 1 \choose n + 1} \end{align}

Felix Marin
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