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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?

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    $\begingroup$ 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 $k$). $\endgroup$
    – Gjergji Zaimi
    Commented Jun 15, 2013 at 15:59
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    $\begingroup$ 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. $\endgroup$
    – Sam Hopkins
    Commented Jun 15, 2013 at 16:54
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    $\begingroup$ 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". $\endgroup$
    – Timothy Chow
    Commented Jun 16, 2013 at 18:03
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    $\begingroup$ 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 $k$). $\endgroup$
    – Todd Trimble
    Commented Apr 9, 2016 at 12:48

2 Answers 2

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You could ask Michael Hardy who is the Wikipedia user responsible for adding this quote of Rota to the web page. By doing a binary search of the history for this article you will find on this history page for the article the entry of 23:23, 26 September 2004, due to Michael Hardy, which was the first entry in which this quote appears.

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    $\begingroup$ I have asked Michael Hardy, who said "I believe in 1998 or '99 he had a paper in the Mathematical Intelligencer that included that. I seem to recall that in three consecutive issues of the Intelligencer there were papers by him, which were lectures he had given at a coference. I think it was in one of those." However, I couldn't find the quote in those articles. $\endgroup$
    – Vince Vatter
    Commented Jun 15, 2013 at 15:00
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In particular, these are considered because of applications. See http://hal.archives-ouvertes.fr/docs/00/04/32/07/PDF/PI-1762.pdf for an application in chemistry.

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