Binomial Coefficients sum [closed]

Question

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)?

It is supposed that $b \le a$.

Answer 1

Here's a combinatorial proof that the expression is $0$. Both sides of the identity count the number of $(b-1)$-subsets of $\{1,\dots,a+b-1\}$ that include $\{1,\dots,b\}$. Because $b > b-1$, this count is obviously $0$, establishing the RHS. For the LHS, apply inclusion-exclusion, where the $b$ properties to be avoided are that $j$ does not appear for $j \in \{1,\dots,b\}$. More generally, this argument shows that $$\sum_{i=0}^b (-1)^i \binom{b}{i} \binom{a+b-1-i}{k} = 0$$ for $k < b$, and it does not require $b \le a$.

Answer 2

This is always zero. Combinatorially, consider $a + b - 1$ balls. Choose $a$ balls, and color them into black and white such that only balls $1, \ldots, b$ can be black. The sum in OP is the difference between the colorings with even and odd numbers of black balls ($i$ being the number of black balls). However, valid colorings of any set $S$ of size $a$ cancel out. Indeed, $B = S \cap \{1, \ldots, b\}$ is not empty, thus the sum is $\sum_{A \subseteq B} (-1)^{|A|} = 0$.

Answer 3

$(1+x)^b=\sum_{k=0}^{b} \binom{b}{k}x^k$

Now, $\binom{a+b-1-i}{a-i}=\binom{a+b-1-i}{b-1}$...$(1)$

And, $(1+x)^{-b}=1-\binom{b}{b-1}x+\binom{b+1}{b-1}x^2-..\color{cadetblue}{(-1)^{a-b}\binom{a-1}{b-1}x^{a-b}+(-1)^{a-b+1}\binom{a}{b-1}x^{a-b+1}......+(-1)^{a}\binom{a+b-1}{b-1}x^{a+1}}+.... \tag{2}$

Multiplying (1) and (2) we easily see that the coefficient of $x^a$ in r.h.s $$(-1)^a\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$.

But the left hand side of $(1)×(2)$ is 1. Hence, the coefficient of $x^a, a\geq 1$ is $0$.

Hence, $$\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}=0$$