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I know the sums $$\sum_{q=0}^{d-r}\sum_{s=r+q}^{d}{{\binom {r-1+q}{r-1}}(r-1)!s}$$ and $$\sum_{s=r}^{d}{\binom{s}{r}(r-1)!s}$$ are equal. How do I manipulate the double sum to look exactly like the single sum?

Mike Earnest
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Poisson
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  • Not sure what you want here. Do you want to show how to get from the double sum to the single sum? – delivery101 Nov 29 '21 at 12:16
  • Yes, please. If there is a way you can do it, please let me know. – Poisson Nov 29 '21 at 13:18
  • It would improve your Question if you were more specific about how you "know the sums ... are equal". One way to show you've digested the problem statement is to verify they are equal for some fixed values $r \le d$. – hardmath Nov 30 '21 at 06:51

1 Answers1

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The first trick is to switch the order of summation, which is often a very fruitful trick in simplifying double combinatorial sums. To do this, we first use an Iverson bracket to write the double sum in a way so the inner sum bounds do not depend on the outer sum variable. The expression $[s\ge r+q]$ is $1$ if the condition $s\ge r+q$ is true, $0$ otherwise. In the third equation, we get rid of the Iverson bracket by modifying the new inner summation, essentially the first trick in reverse.

$$ \begin{align} \sum_{q=0}^{d-r}\sum_{s=r+q}^{d}{{\binom {r-1+q}{r-1}}(r-1)!s} &=\sum_{q=0}^{d-r}\sum_{s=r}^{d}{{\binom {r-1+q}{r-1}}(r-1)!s}[s\ge r+q]\\ &\hspace{1cm}\searrow\hspace{-.5cm}\swarrow\\ &=\sum_{s=r}^{d}\sum_{q=0}^{d-r}{{\binom {r-1+q}{r-1}}(r-1)!s}[s\ge r+q]\\ &=\sum_{s=r}^{d}\sum_{q=0}^{\color{blue}{s-r}}{{\binom {r-1+q}{r-1}}(r-1)!s}\\ &=\sum_{s=r}^{d}(r-1)!s\sum_{q=0}^{s-r}{{\binom {r-1+q}{r-1}}} \end{align} $$ Finally, the inner summation $\sum_{q=0}^{s-r}{{\binom {r-1+q}{r-1}}}$ simplifies to $\binom{s}r$ using the hockey stick identity.

Mike Earnest
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