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I'd like to know how to prove the following property of binomial coefficientp.169 of Concrete Mathematics, by Ronald Graham, Donald Knuth, and Oren Patashnik $$\sum_{0\leq k\leq l}\binom{l-k}{m}\binom{q+k}{n}=\binom{l+q+1}{m+n+1}$$ There should be a similar method with the proof of Vandermonde's convolution, I understand its 3 proofs on wikipedia but I cannot figure out the proof of this one.

Junya
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  • thanks, these posts help a lot – Junya Jun 23 '24 at 23:45
  • proof in the link is excellent, but I don't quite understand how is $n\geq r$ is relevant. – Junya Jun 23 '24 at 23:58
  • sorry for the unclarity, I mean $n \geq q \geq 0$ in my question. – Junya Jun 24 '24 at 00:44
  • My quick guess is to consider only when the second factor starts from $\binom{q+0}n = 0$ or $=1$, and not miss any terms. For example, $$\sum_{k=0}^2 \binom{2-k}{2}\binom{3+k}{1} = 1\cdot 3+0\cdot 4+0\cdot 5 = 3\ \binom{2+3+1}{2+1+1} = 15$$ because the LHS is missing $k=-2,-1$: $$\sum_{k=-2}^{-1} \binom{2-k}{2}\binom{3+k}{1} = 6\cdot 1+3\cdot 2 = 12$$ – peterwhy Jun 24 '24 at 01:26
  • I understand it in examples, but as I read the proof it uses Vandermonde's identity, upper negation, symmetry. I don't know which one is only true for non-negative numbers? in other words, if the condition $n\geq q\geq 0$ is removed, if we write the same proof we need to notice the need to discuss cases and separate terms at a certain step, and which step is it? – Junya Jun 24 '24 at 03:09
  • If you are referring to this answer under the duplicate, I suspect it's about the particular form of Vandermonde's:

    $$\sum_{k=\color{red}0}^b\binom a{b-k}\binom c{d+k} = \binom{a+c}{b+d}$$

    The LHS may be missing some of the full $b+d$ terms if $d>0$, i.e. if $q-n>0$ here. (But I am too tired to consider all cases, including negative binomial coefficients, now)

     – peterwhy Jun 24 '24 at 04:36

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