Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
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. –  Henning Makholm Apr 24 '12 at 17:27
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. –  Thomas Andrews Apr 24 '12 at 17:28
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". –  Michael Hardy Apr 24 '12 at 17:36
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... –  J. M. Apr 24 '12 at 17:51
add comment

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.