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 $\binom{4}{-3}$?

  • $\begingroup$ You don't have any "result" to explain yet. So far you have just defined a meaning for some notation that didn't have meaning before. There's no result to explain until you have a claim about how that notation can be used. $\endgroup$ – hmakholm left over Monica Apr 24 '12 at 17:27
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    $\begingroup$ Why must there be a combinatorial meaning? I mean, the sum $1+2+3+...+n = \frac{n(n+1)}2$ when $n$ is positive, but does it make sense to give this formula a meaning when $n<0$? What about when $n=\frac{1}{2}$? There are usages for the negative combinations (and fractional combinations), related to the power series for $(x+1)^n$ when $n$ is negative or not an integer. $\endgroup$ – Thomas Andrews Apr 24 '12 at 17:28
  • $\begingroup$ A can't answer this, but $20$ is the number of multisets of size $3$ whose members are those of a set of size $4$, and if I recall correctly $\left|\dbinom{-n}{k}\right|$ is the number of multisets of size $k$ whose members are those of a set of size $n$, for $n,k\in\{0,1,2,3,\ldots\}$. Google the term "multiset coefficients". $\endgroup$ – Michael Hardy Apr 24 '12 at 17:36
  • $\begingroup$ One way to look at $\binom{4}{-3}$ would be to consider the interpretation of $(1+x)^\alpha=\sum\limits_{k=-\infty}^\infty \binom{\alpha}{k}x^k$... notice how the binomial series doesn't have negative powers, which means the corresponding coefficients are zero, which means... $\endgroup$ – J. M. is a poor mathematician Apr 24 '12 at 17:51

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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