<s>[INST] Combinatorics sum to 1 using Identity I want to prove that the following equals 1:
$W=\sum_{i=0}^{n-r}(-1)^i n \frac{C_{n-1}^{r-1}C_{n-r}^i}{(r+i)}$. I tried mathematical induction and succeeded. Is there any known identity of combinatorics that could be used to prove it directly?
 [/INST] A Slightly More General Identity
$$
\begin{align}
\sum_k(-1)^k\frac{\binom{n}{k}}{\binom{r+k}{m}}
&=\sum_k(-1)^k\frac{\binom{n-1}{k}+\binom{n-1}{k-1}}{\binom{r+k}{m}}\tag1\\
&=\sum_k(-1)^k\binom{n-1}{k}\left(\frac1{\binom{r+k}{m}}-\frac1{\binom{r+k+1}{m}}\right)\tag2\\
&=\sum_k(-1)^k\binom{n-1}{k}\left(\frac{\frac{r+k+1}{m+1}}{\binom{r+k+1}{m+1}}-\frac{\frac{r+k+1-m}{m+1}}{\binom{r+k+1}{m+1}}\right)\tag3\\
&=\frac{m}{m+1}\sum_k(-1)^k\frac{\binom{n-1}{k}}{\binom{r+k+1}{m+1}}\tag4\\
&=\frac{m}{m+n}\frac1{\binom{r+n}{m+n}}\tag5
\end{align}
$$
Explanation:
$(1)$: Pascal's Rule
$(2)$: substitute $k\mapsto k+1$ in $(-1)^k\left.\binom{n-1}{k-1}\middle/\binom{r+k}{m}\right.$
$(3)$: put things over a common denominator
$(4)$: subtract and pull out the factor of $\frac{m}{m+1}$
$(5)$: repeat $n$ times

Application to the Question
$$
\begin{align}
\sum_{k=0}^{n-r}(-1)^kn\frac{\binom{n-1}{r-1}\binom{n-r}k}{r+k}
&=n\binom{n-1}{r-1}\sum_{k=0}^{n-r}(-1)^k\frac{\binom{n-r}k}{\binom{r+k}{1}}\\
&=n\binom{n-1}{r-1}\frac1{n-r+1}\frac1{\binom{n}{n-r+1}}\\
&=n\binom{n-1}{r-1}\frac1n\frac1{\binom{n-1}{n-r}}\\[6pt]
&=1
\end{align}
$$
</s>