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Any idea on whether or not $$\sum_{i = 0}^a(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$$$\sum_{i = 0}^b(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$
has a closed formula on $a$ and $b$ (and on what it is, in case it does)?
It is supposed that $b \le a$.
Any idea on whether or not $$\sum_{i = 0}^a(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$
has a closed formula on $a$ and $b$ (and on what it is, in case it does)?
Any idea on whether or not $$\sum_{i = 0}^b(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$
has a closed formula on $a$ and $b$ (and on what it is, in case it does)?
Any idea on whether or not $$\sum_{i = 0}^a(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$
has a closed formula on $a$ and $b$ (and on what it is, in case it does)?
