Are binomial coefficients defined for $k \in \mathbb{R}$

Question

So I've been trying to learn some basic precalc and I got stumped on definition of combinations namely the following confuses me .

I know that The symbol ${n\choose k}$ is read as "$n$ choose $k$." It represents the number of ways to choose $k$ objects from a set of $n$ objects. However in most definitions no restriction on the value of k is made for example what would ${5\choose 2.5}$ be.

furthermore what do we define ${n\choose k}$ if $k<0$

Some help would be greatly appreciated.

Thanks in advance.

Answer 1

The function $\binom{n}{k}$ can be extended to all real (and even complex) $n,k$ by using the Gamma function in place of factorials: $$ \binom{n}{k}:=\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)} $$ This suffices to define $\binom{n}k$ except when $k$ is a negative integer or $n-k$ is a negative integer, since $\Gamma$ has a pole at all non-positive integers. We can fix some of these exceptional cases by using a limit: $$ \binom{n}{k}:=\lim_{(x,y)\to (n,k)}\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)} $$ It turns out that this limit exists, and is zero, if at least one of the following is true:

• $k$ is a negative integer, and $n$ is not a negative integer.

• $n-k$ is a negative integer, and $n$ is not a negative integer.

• $k$ and $n-k$ are both negative integers.

This only leaves the following two cases, where $\binom{n}k$ has a non-removable discontinuity:

1. $k$ and $n$ are negative integers, but $n-k$ is a nonnegative integer.

2. $n-k$ and $n$ are negative integers, but $k$ is a nonnegative integer.

In these cases, we can instead define $\binom{n}{k}$ to be the limit as $(x,y)\to (n,k)$ along a path of a certain slope. That is, we pick a certain real number $\theta$, and define $$ \def\ep{\varepsilon} \binom{n}{k} :=\lim_{\ep\to 0}\frac{\Gamma(n+1+\ep)}{\Gamma(k+1+\theta \ep)\Gamma(n-k+1+(1-\theta)\ep)} $$ Any choice of $\theta$ leads to a valid extension of the binomial coefficients to the entire real plane with nice properties. Namely, Pascal's rule $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$ is now true for all real $n$ and $k$.

• The usual choice is $\theta=0$. Then $\binom{n}k$ is zero in case $1$, and is $\frac{n(n-1)\cdots(n-k+1)}{k!}$ in case $2$. With this convention, we have the nice property that $(1+x)^n=\sum_{k= 0}^\infty\binom{n}k x^k$ holds for all complex $n$.

• Another nice choice is $\theta=1$, where $\binom{n}k$ is $\frac{n(n-1)\cdots(k+1)}{(n-k)!}$ in case $1$, and is zero in case $2$.

• The only choice of $\theta$ which preserves the symmetry property $\binom{n}k=\binom{n}{n-k}$ for all real $n,k$ is $\theta=\frac12$, in which case $\binom{n}k$ is the average of the two previous definitions. In general, different values of $\theta$ lead to weighted averages of the last two cases.

Answer 2

The definition of the factorial is commonly extended to reals by means of the Gamma function. The application to the Binomial coefficients is obvious, and directly relates to the Beta integral.

https://en.wikipedia.org/wiki/Beta_function#Properties

The values at integer $k<0$ and $k>n$ are usually defined to be zero, and if I am right, this is consistent with the definition in terms of Gamma.