I came across Negative Numbers in Combinatorics while working on STA 732 HW1. It’s just one technique needed (after googling) to get me through the problem set, but somehow I find it far more interesting than the problem set (with all due respect to 732).

[Updated 20210514] I started this on the date of HW1 submission, and now I’m finishing it up in May after the whole semester has ended, and that I have finally managed to change the setup from KaTex to MathJax …

Suppose a set S has n elements, number of subsets of S of size k is:

\begin{equation*} {n \choose k} = \frac{n(n-1) \dots (n-k+1)}{k!}. \end{equation*}

Now if there were a set with -2 number of element, how many subsets of size 3 would it have? We can try naively applying the same formula:

\begin{equation*} {-2 \choose 3} = \frac{(-2)(-3)(-4)}{3!} = -4. \end{equation*}

This gives us a number, but how to understand it?

Note that a set, by definition can only consist of unique elements. here we introduce the notion of multisets, with unbounded multiplicity. For example, the following are three different and yet valid multisets: {x,x}, {x,y}, and {y,y}.

Now coming back to the question above, let S = {x,y}, where both x and y are “negative elements”, i.e., they can be selected more than once into a multiset. Then the unique subsets of S of size 3 are: {x,x,x}, {y,y,y}, {x,x,y}, {x,y,y}, a total of 4, which matches with the number 4 we got above. But what about the negative sign? Here let’s say that a multiset with k negative elements has weight of \((-1)^k\). And hence we have

Theorem: For a set S with m negative elements, \({-m \choose k}\) is the sum of the weights of all the k-element multiset-subsets of S.


\[\begin{aligned} {-m \choose k} &= \frac{(-m)(-m-1) \dots (-m-k+1)}{k!} \\ &= (-1)^k {m+k-1 \choose k}. \end{aligned}\]

Now we need to show that the number of unique multisets by choosing k from m elements, is indeed \({m+k-1 \choose k}\). We can use a counting method called “stars and bars”. This might be rather basic, but it’s the first time I come across this idea. I find it quite interesting.

Now think about choosing k from 2 negative elements. As the order doesn’t matter here, this is equivalent to … imagine we have k stars, lying in one line

\[* * * \dots * * *\]

We want to put one bar in between to separate these stars into two groups, for example this is one possibility:

\[* | * * \dots * * *\]

How many different ways are there to place this bar? This is exactly equivalent to the problem above. Here there are a total of (k+1) places for me to place this bar, and hence it’s \({k+1 \choose 1} = {2+k-1 \choose 2-1}\) number of different possibilities. And if we generalize it an (m choose k) problem, obviously the answer is

\[{m+k-1 \choose m-1} = {m+k-1 \choose k}.\]

This completes the proof.

This also works for hybrid sets, i.e., sets with both positive and negative elements:

  • a positive element can only be chosen once, with weight 1
  • a negative element can be chosen any number of times, with weight (-1)

For example, consider a set S = {a, x, y} where a is a positive element, while x and y are negative elements. In total S has (-1) elements. So how many subsets (potentially multisets) of size 3 does S have?

  1. For subsets with only negative elements, each subset/ multiset has weight of \((-1)^3 = (-1)\) and there are a total of 4 subsets
  2. For subsets with one positive element a, and two negative elements, the weight is 1, and there are a total of 3 subsets.
  3. All together, the sum of the weights of all the unique subsets is 3-4=-1. This is same as \(-1 \choose 3\).

This is where I decided to stop. I did not try to further prove the case for hybrid sets.


  • For general understanding of this concept: http://faculty.uml.edu/jpropp/msri-up12.pdf
  • For the “stars and bars” technique: https://brilliant.org/wiki/integer-equations-star-and-bars/