Given that $$\sum_{n=2}^8 \frac{1}{n!(17-n)!} = \frac{N}{16!},$$ find the greatest integer less than $\frac{N}{100}.$
Writing out the summation, I noted that the hockey stick identity could potentially be put to use here, but I was unable to proceed from there.
Hint: using a symmetry in $\frac{1}{n!(17-n)!}$ you can write: $$\sum_{n=0}^8 \frac{1}{n!(17-n)!} = \frac{1}{2} \sum_{n=0}^{17} \frac{1}{n!(17-n)!} = \frac{1}{2} \cdot \frac{1}{17!} \sum_{n=0}^{17} \binom{17}{n} = \frac{1}{2} \cdot \frac{1}{17!} \cdot 2^{17}.$$ Now you just have to see how to make the adjustment to get an expression for $\sum_{n=2}^8 \frac{1}{n!(17-n)!}$ instead.