Intuition behind negative combinations
Take $\binom{n}{r}$. It denotes how in how many different ways you can choose $r$ elements from a set of $k$ elements. For case $\binom{4}{3}$ which evaluates to $\frac{4!}{3!(4-3)!}=4$, it perfectly makes sense.
However, consider $\binom{-4}{3}$. Evaluating it by factoring out $(n-r)!$ from the numerator and the denominator, we get $\frac{(-4)_{3}}{3!}$, where $(-4)_{3}$ is a falling factorial. Hence, $\binom{-4}{3}=-20$.
Now, how do we explain this result? Following the combinatoric reasoning, how can there be a negative number of ways to choose $r$ elements from a set containing a negative number of elements? And if counting with negative ways and negative elements is possible, why can't we check how many negative ways there are of choosing $r$ elements, for example $\binom{4}{-3}$?
$\endgroup$ 41 Answer
$\begingroup$I know two "explanations" of this phenomenon. One idea is to replace cardinality with a form of Euler characteristic; this is described, for example, in Propp's Exponentiation and Euler measure.
The other is to replace cardinality with dimension (of a vector space) and then come up with a reasonable notion of negative dimension. The starting observation is that if $V$ is a vector space of dimension $n$, then the exterior power $\Lambda^k(V)$ has dimension ${n \choose k}$, whereas the symmetric power $S^k(V)$ has dimension $(-1)^k {-n \choose k}$. It turns out one can think of the exterior power as being the symmetric power but applied to a "vector space of negative dimension." I briefly explain this story in this blog post.
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