Why are negative sets multisets? (Reference request)

It is easy to establish that $$\left(\!\!{n\choose k}\!\!\right)=(-1)^k{-n \choose k},$$ where the symbol on the left-hand-side counts the number of multisets of $k$ elements from $n$.

On the Wikipedia page for multisets, it is further claimed that "This fact led Gian-Carlo Rota to ask "Why are negative sets multisets?". He considered that question worthy of the attention of philosophers of mathematics."

While I think it is quite plausible that Rota may well have asked this question, no citation is provided for it on Wikipedia, and my attempts to source the quote have all proved fruitless. Have you seen a quote by Rota relating "negative sets" to multisets?

• I can't resist pointing out that this interpretation of "negative sets are multisets" is made very clear by the "Euler characteristic as generalized cardinality" point of view. In fact, since we know that $\chi (Sym^k X)=\binom{\chi(X)}{k}$, we notice that $\binom{n}{k}$ counts the number of ways of picking $k$ points among $n$ points, but on the other hand, $(-1)^k\binom{-n}{k}$ counts the number of ways of picking $k$ points among $n$ intervals (this is the same as multisets of size $k$). – Gjergji Zaimi Jun 15 '13 at 15:59
• This paper by D. Loeb, "Sets with a negative number of elements": faculty.uml.edu/jpropp/negative.pdf, has a reference to a paper that Loeb and Rota wrote together. Perhaps there is a clue there- I can't find it online, though. – Sam Hopkins Jun 15 '13 at 16:54
• I don't know the answer, but if you have not done so already, I'd recommend that you check the books "Discrete Thoughts" and "Indiscrete Thoughts". – Timothy Chow Jun 16 '13 at 18:03
• On the other hand, I'm very tempted to view this in terms of superalgebra, interpreting $\binom{n}{k}$ as [the dimension of] the $k$-th alternating power of an $n$-dimensional vector space in ordinary algebra, but when we apply the functorial construction to an $n$-dimensional super space concentrated in the odd component, we get a $k$-th symmetric power of the underlying space (concentrated in the component of parity $k$). – Todd Trimble Apr 9 '16 at 12:48