Prove that $\sum_{q=0}^{d-r}\sum_{s=r+q}^{d}{{\binom {r-1+q}{r-1}}(r-1)!s}=\sum_{s=r}^{d}{\binom{s}{r}(r-1)!s}$ by sum manipulation 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?
 A: 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.
