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Mar 10, 2022 at 13:45 history edited Vince Vatter CC BY-SA 4.0
Put quote in quote environment
Apr 9, 2016 at 12:48 comment added Todd Trimble 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 $k$).
Apr 9, 2016 at 0:51 history edited Vince Vatter CC BY-SA 3.0
Mixed multiset typesetting
Jun 16, 2013 at 18:03 comment added Timothy Chow 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".
Jun 16, 2013 at 6:16 answer added Mark timeline score: -2
Jun 15, 2013 at 16:54 comment added Sam Hopkins 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.
Jun 15, 2013 at 15:59 comment added Gjergji Zaimi 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 $k$).
Jun 15, 2013 at 13:59 answer added Lee Mosher timeline score: 6
Jun 15, 2013 at 8:25 history asked Vince Vatter CC BY-SA 3.0