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.

  • $\begingroup$ 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? $\endgroup$ – shuhalo Sep 15 '20 at 22:06
  • $\begingroup$ @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. $\endgroup$ – Qiaochu Yuan Sep 15 '20 at 22:15

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