For a binomial coefficient $$\binom ab$$ would it be correct to say the following:

  1. $b$ must be either $0$ or a positive integer. i.e. $b$ cannot be negative or a fraction.

  2. $a$ can be either positive or negative, and either an integer or a fraction, subject to the condition that if $a$ is a positive integer, then $a\ge b$ (otherwise the binomial coefficient is defined as zero).

This means that we can have binomial coefficients like $$\binom {-2}3=\frac {(-2)(-3)(-4)}{1\cdot 2\cdot 3}$$ $$\binom {-\frac 13}4=\frac {-\frac 13\cdot -\frac 43\cdot -\frac 73\cdot -\frac {11}3}{1\cdot 2\cdot 3\cdot 4}$$ But binomial coefficients like $$\binom 34=0$$ as $3<4 (3,4\in \Bbb{Z})$ whilst $$\binom {3}{\frac 14}$$ is not defined.

Are there any other conditions which have not been included? Does a binomial coefficient exist for numbers which are not rational?

[Note - following from comments on this question, it appears that the limitations on parameters of a binomial coefficient $$\binom ab$$ are that both $a,b$ are real.

If follows from the same definition that if $a$ is an integer less than $b$ then then $\binom ab=0$, because of the "zero crossing" in the falling factorial of $a$.]

  • $\begingroup$ The upper number, $a$, can be any real number; the lower number, $b$, can be any non-negative integer. There is no requirement that $a\ge b$ when $a$ is a positive integer; as you say, $\binom34$ is defined and is $0$. $\endgroup$ – Brian M. Scott Oct 31 '16 at 14:37
  • $\begingroup$ You can extend the definition of the binomial coefficients to reals by using the extension of the factorial based on the Gamma function. Among others, this leads to the generalized binomial theorem for real powers $(a+b)^r$. $\endgroup$ – Yves Daoust Oct 31 '16 at 14:38
  • $\begingroup$ @BrianM.Scott Thanks. That's a good point. It's zero because, following the same definition, $\frac {3\cdot 2\cdot 1\cdot 0}{1\cdot 2\cdot 3\cdot 4}=0$, which is not the same ans not being defined. $\endgroup$ – hypergeometric Oct 31 '16 at 14:39
  • $\begingroup$ @hypergeometric: Exactly. And of course that also agrees with the definition (for $a,b\in\Bbb N$) that $\binom{a}b$ is the number of $b$-element subsets of $[a]$. $\endgroup$ – Brian M. Scott Oct 31 '16 at 14:40
  • $\begingroup$ @hypergeometric: or because $3!/4!(-1)!=0$. $\endgroup$ – Yves Daoust Oct 31 '16 at 14:40

From $r!:=\Gamma(r+1)$, you can define

$$\binom rs:=\frac{r!}{(r-s)!s!}$$

for any reals (but negative integers) and when either factor at the denominator is a negative integer, the expression is defined as $0$.

For instance,

$$\binom rk=\frac{r!}{(r-k)!k!}=\frac{\Gamma(r+1)}{\Gamma(r-k+1)k!}=\frac{r(r-1)\cdots(r-k+1)}{k!}=\frac{(r)_k}{k!}$$ where the numerator is a so-called falling factorial.

This allows to write the generalized binomial theorem as

$$(a+b)^r=\sum_{k=0}^\infty\binom rka^{r-k}b^k.$$

  • $\begingroup$ Just to confirm, the $a,b$ in the first equation are not the same as $a,b$ in the last equation. Is that right? $\endgroup$ – hypergeometric Oct 31 '16 at 15:04
  • 1
    $\begingroup$ @hypergeometric: now rewritten. $\endgroup$ – Yves Daoust Oct 31 '16 at 15:07

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