According to various sources e.g.


and Wolfram themselves

http://functions.wolfram.com/GammaBetaErf/Binomial/02/ , the binomial coefficient ${n\choose k}$ is is defined as 0 whenever $k$ and $n$ are negative integers and $k\le n$. But when I type


Mathematica returns


I looked up the documentation for the definition of Binomial and it says

In general, ${n\choose m}$ is defined by $\Gamma(n+1)/\Gamma(m+1)\Gamma(n-m+1)$ or suitable limits of this.

Apparently, when $n=-1$ or $m=-1$ since $\Gamma(0)$ is not defined the suitable limit case is applied.

So, why does Mathematica return 1 for ${-1\choose -1}$? What precisely is the "suitable limits"?


2 Answers 2


Looked up the rules on this. This is how it works. If $n$ and $r$ are negative integers, there is a symmetry relation $\binom{n}{r}=\binom{n}{n-r}$ and now the limit is used.

Mathematica graphics

Mathematica graphics

But now $\binom{n}{r}=\binom{-1}{0}$ from above. Hence the above limit is, where $n=-1$ and $r=0$ is

    n = -1; r = 0;
    Limit[ Gamma[n + t + 1]/(Gamma[r + 1] Gamma[n + t - r + 1]), t -> 0]
    (* 1 *)

Maple also agrees with Mathematica

Mathematica graphics

Reference: Maple help on Binomial

  • $\begingroup$ Thanks. I've never used Maple. Do you mean Mathematica uses the same definition as Maple? $\endgroup$
    – Y. Pei
    Commented Apr 10, 2014 at 20:51
  • $\begingroup$ Also, the documentation of Maple is not clear either. It says "In the case that n is a negative integer, binomial(n,r) is defined by this limit." Note here's a full stop at the end of the sentence. So binomial(-1,-1) should be using this limit $\lim_{t\to0}{\Gamma(-1+t+1)\over \Gamma(-1+1)\Gamma(-1+t-(-1)+1)}$, which is ill-defined because of the $\Gamma(0)$ in the denominator. Even if when k is a negative integer we use the symmetry property, then the definition does not cover binomial(-3,-2) properly because then binomial(-3,-2) turns into binomial(-3,-1), which is not well-defined. $\endgroup$
    – Y. Pei
    Commented Apr 10, 2014 at 21:02
  • $\begingroup$ Result from PARI (version 2.7.2): binomial(-1,-1) = 0 $\endgroup$ Commented Jun 13, 2015 at 21:25

Without expanding Binomial into Gamma functions, you can also see that the result is correct based on the following true statement:

SeriesCoefficient[(1 + x)^n, {x, 0, k}, Assumptions -> k >= 0]

(* ==> Binomial[n, k] *)

This is the binomial expansion, valid in particular for n = -1. But that case leads to the alternative expression

SeriesCoefficient[(1 + x)^-1, {x, 0, k}, 
 Assumptions -> k >= 0]

(* ==> (-1)^k *)

which by comparison with the previous line implies that the following is the correct result:

Binomial[-1, 0]

(* ==> 1 *)

Now we have the symmetry relation

FullSimplify[Binomial[n, k] == Binomial[n, n - k]]

(* ==> True *)

Using this with n = -1 and k = -1 then confirms what Mathematica says:

Binomial[-1, 0] == Binomial[-1, -1]

(* ==> True *)

Combined with the previous equation the result Binomial[-1, -1] == 1 follows.


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