Someone submitted a new entry (entitled "Don Knuth Plays with ChatGPT") to Hacker News a few hours ago. The entry pointed to a text file that documented twenty questions posed to ChatGPT, the answers returned as well as Donald Knuth's comments. Question 3 is as follows:
Why does Mathematica give the wrong value for Binomial[-1,-1]?
ChatGPT answered that
The expression Binomial[-1,-1] is undefined, and so there is no unique or "correct" value that can be assigned to it. ... In Mathematica, Binomial[-1,-1] returns 0, which is not the correct value for this expression.
Knuth commented that (emphasis mine):
Answer #3 is fouled up beautifully! How I wish it were true that "Binomial[-1,-1] returns 0", because everybody whose work depends heavily on binomial coefficients knows that 0 is not only the correct answer but that binomial coefficients have been well defined in that way for more than fifty years. And of course, unfortunately, Binomial[-1,-1] actually returns 1, alas.
The answer goes on to state that (-1)! is well defined (indeed equal to 1), etc. But it's all expressed in marvelous English prose.
Why should $\binom{-1}{-1}$ be defined as zero? Is it a convenient convention (if so, why is it convenient?) or is there a proper generalization of $\binom{n}{k}$ to negative arguments that evaluates to $0$ when $n=k=-1$? I thought that $$\binom{n}{k}=\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}$$ when both the numerator and denominator are well-defined. When $n=k=-1$, although the gamma function is not defined at zero, if we cancel out the two instances of $\Gamma(0)$ in the numerator and denominator, isn't it more sensible to define $\binom{-1}{-1}$ as $\frac{1}{\Gamma(1)}=1$?
For combinatorics, the appropriate context to talk about $\binom{n}{k}$ is generating functions, not meromorphic functions. This shouldn't be surprising that there are different ways to extend a concept.
In this context, $\binom{n}{k}$ is the coefficient of $z^k$ of $(1+z)^n \in \mathbb{R}[[x]]$.
Here, $n$ may be any real number, but definitionally any $k$ which isn't a non-negative integer would result in $0$.