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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 $a$.]

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  • $\begingroup$ 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
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    $\begingroup$ 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. $\endgroup$ – Hypergeometricx Oct 31 '16 at 14:39
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    $\begingroup$ @hypergeometric: or because $3!/4!(-1)!=0$. $\endgroup$ – Yves Daoust Oct 31 '16 at 14:40
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    $\begingroup$ No, $b$ can also be a real number. $\binom{3.14}{2.72}=1.89557\cdots$. $\endgroup$ – Yves Daoust Oct 31 '16 at 15:05
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$$ \binom {-2}3=\frac {(-2)(-3)(-4)}{1\cdot 2\cdot 3} $$ $$ \binom {-\frac 13}4=\frac {(-\frac 13)(-\frac 43)(-\frac 73)(-\frac {11}3)}{1\cdot 2\cdot 3\cdot 4} $$ Things like this appear in such infinite series as $$ (x+y)^{-1/3} = \sum_{k=0}^\infty \binom {-\frac 1 3} k x^k y^{-1/3-k} $$ which converges to the expression on the left if $\left| \dfrac x y \right| <1 $ and otherwise diverges.

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  • $\begingroup$ I'm grateful that you also provided an answer here, your kindness makes this forum better. $\endgroup$ – linear_combinatori_probabi Sep 5 '20 at 12:38
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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

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

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