# What are the domain and values of binomial coefficients $n \choose k$ for any integer $n$ and $k$, and why?

This is a question about "nasty details" of the binomial coefficients.

I would like to understand the definition of binomial coefficients $$n \choose k$$ for general integers $$n$$ and $$k$$.

One way to define binomial coefficients is as the number of cardinality $$k$$ subsets of an cardinality $$n$$ subsets, which is,

$${n \choose k} := |\{\; S \subseteq \{ 1, \dots, n \} \;:\; |S| = k \;\}|$$

This formula is well-defined for any integer $$n$$ and $$k$$. Note that for $$k = 0$$ we always get $${n \choose k} = 1$$ regardless of $$n$$ but generally it is zero for negative $$n$$.

There are many other ways of defining the binomial coefficients. For instance, another definition is:

$${n \choose k} := \prod_{i=1}^k \dfrac{ n+1-i }{ i }$$

which is equals $$1$$ for any non-positive $$k$$ regardless of $$n$$. Argueably, the latter definition of the binomial coefficients is not regarded as "foundational" for non-negative $$n$$ and $$k$$, but that does not really help in deciding for integers the binomial coefficients are defined and what their values are.

Is there a commonly accepted definition of $${n \choose k}$$ for any integers $$n$$ and $$k$$, and what are the benefits of that definition for the working mathematician?

My preferred definition is:

$$\binom{x}k=\begin{cases} \frac{x^{\underline{k}}}{k!},&\text{if }0\le k\in\Bbb Z\\ 0,&\text{if }0>k\in\Bbb Z\,, \end{cases}$$

where $$x$$ can in principle be any complex number (though I’ve only actually seen it used with $$x\in\Bbb R$$), and $$x^{\underline{k}}$$ is a falling factorial. This behaves correctly for non-negative integer values of $$x$$ and $$k$$, behaves as it ought for negative integer $$k$$, works well in connection with manipulation of generating functions, and makes the binomial coefficient a polynomial in $$x$$ of degree $$k$$, which can be useful.

If you want the binomial coefficients $${s \choose k}$$ to satisfy the binomial theorem

$$(1 + x)^s = \sum_{k \ge 0} {s \choose k} x^k$$

in the greatest generality possible, then by repeatedly taking derivatives you can see that you are required to define

$$\boxed{ {s \choose k} = \frac{s(s-1) \dots (s - (k-1))}{k!} }.$$

Here $$k$$ is still a nonnegative integer but $$s$$ can be an arbitrary complex number (at least; $$s$$ can take values in any commutative $$\mathbb{Q}$$-algebra). This definition together with the binomial theorem shows that, for example, we still have Vandermonde's identity

$${s+t \choose k} = \sum_{i+j=k} {s \choose i} {t \choose j}$$

for arbitrary complex $$s, t$$, and in fact as a polynomial identity in $$s$$ and $$t$$.

Specializing, if $$s$$ is a negative integer we get the negative binomial coefficients, which are combinatorially meaningful since they describe the Taylor series expansion of

$$\frac{1}{(1 - x)^n} = \sum_{k \ge 0} (-1)^k {-n \choose k} x^k$$

which gives that $$(-1)^k {-n \choose k} = {n+k-1 \choose k}$$ is the number of solutions to $$a_1 + \dots + a_n = k$$ for non-negative integers $$a_i$$; see stars and bars for more on this. See also, for example, the negative binomial distribution.

Even non-integer values of $$s$$ are combinatorially meaningful; for example $$s = -\frac{1}{2}$$ shows up in the generating function of the central binomial coefficients. If we consider $$s$$ to be a formal variable then $$\frac{1}{(1 - x)^s}$$ can be thought of as a two-variable generating function for the (unsigned) Stirling numbers of the first kind (and we get the signed Stirling numbers of the first kind with $$(1 + x)^s$$).

Some people might go further and generalize $$k$$ using the Gamma function but I have personally never needed to do this. I know exactly one place where it shows up, which is the Beta function. My preferred convention is that $${s \choose k}$$ is only defined for $$k$$ a nonnegative integer; that's all I've ever needed.

• I know at least one reason to define $n \choose k$ for negative $k$: the alternating algebra for $\bigwedge^k \mathbb R^n$ should be zero-dimensional for negative $k$, so that the de Rham complex has trivial infinite extensions in both directions. That motivates setting ${n \choose k} := 0$ for negative $k$ and non-negative $n$. What do you think? – shuhalo Sep 15 '20 at 22:06
• @shuhalo: I guess it’s a matter of convention but I would just say that it’s undefined. The exterior power is a certain quotient of the tensor power and the tensor power is undefined for negative exponent unless the input is invertible. – Qiaochu Yuan Sep 15 '20 at 22:15