<s>[INST] How do you evaluate this summation: $S=\sum\limits_{r=0}^{15} (-1)^r \frac{\binom{15}{r}}{\binom{r+3}{r}}$ Find S:
$$S=\sum_{r=0}^{15} (-1)^r \frac{\binom{15}{r}}{\binom{r+3}{r}}$$

My attempt:
I tried writing the summation as:
$$S=3!(15!)\sum_{r=0}^{15} (-1)^r \frac{1}{(15-r)!(r+3)!}$$
and tried to convert it to a telescoping form, but to no avail.

I don't know how to go about solving this question and any help will be appreciated.
 [/INST] $$S=\sum_{r=0}^{15} (-1)^r \frac{\binom{15}{r}}{\binom{r+3}{r}} = \sum_{r=0}^{15} (-1)^r \frac{15!}{r! (15-r)!} \frac{3!\cdot r!}{(r+3)!}$$
$$= 15!\cdot3!\cdot\sum_{m=3}^{18} (-1)^{m+1} \frac{1}{(18-m)!} \frac{1}{(m)!}$$
$$= \frac{3!}{16 \cdot 17 \cdot 18} \sum_{m=3}^{18} (-1)^{m+1} \binom{18}{m}$$
$$= \frac{3!}{16 \cdot 17 \cdot 18} \left( \left(- \sum_{m=0}^{18} (-1)^{m} \binom{18}{m} \right) + 1 - \binom{18}{1} + \binom{18}{2} \right) $$
Now since
$$\sum_{i=0}^n (-1)^i \binom{n}{i} = 0 $$
it becomes
$$= \frac{3!}{16 \cdot 17 \cdot 18} \left( 1 - \binom{18}{1} + \binom{18}{2} \right) = \frac{1}{6}$$
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