\begin{align} \sum_{k=0}^m (-1)^k \binom{u}{k} \binom{v}{m-k} \tag{1}\\ &= \sum_{k=0}^m (-1)^k\binom{u}{k}[z^{m-k}](1+z)^{v} \tag{2}\\ &= [z^m](1+z)^v \sum_{k=0}^m (-1)^k [z^{-k}]\binom{u}{k} \tag{3}\\ &=[z^m](1+z)^v \sum_{k=0}^m (-1)^k [z^{-k}][z^k](1+z)^u \tag{4}\\ &=[z^m](1+z)^v \sum_{k=0}^m (-1)^k \tag{5}\\ &=[z^m](1+z)^v = \binom{v}{m} \tag{6} \end{align}
if m is even. (5) is zero otherwise. Here $[z^m]$ denotes the coefficent of $z^m$ when $(1+z)^v$ expanded. however if the steps above right, the result is zero if $m > v$. But in the summed notation there can still be some results. another way: $\displaylines{ (1-x)^u (1-x)^v = (1-x)^{u+v} \\ &\Big( \sum_{i =0}^u (-1)^i x^i \binom{u}{i}\Big)* \Big(\sum_{j =0}^v (-1)^j x^j \binom{v}{j}\Big) = \sum_{k =0}^{u+v} (-1)^k x^k \binom{u+v}{k}\\ &\sum_{k =0}^{u+v} \Big(\sum_{i+j=k} \binom{u}{i} \binom{v}{j}\Big)(-1)^k x^k = \sum_{k=0}^{u+v}(-1)^k x^k \binom{u+v}{k} }$ thus $\Big(\sum_{i+j=k} \binom{u}{i} \binom{v}{j}\Big)(-1)^k = (-1)^k \binom{u+v}{k} $ two different results. where do I do mistakes?
