How do you calculate combinations with fractions (eg, "$\frac13$ choose $2$")? I was given "$\frac13$ choose $2$" and asked to compute. I checked the answer and it's $-\frac19$, but I have no clue how to arrive at this answer.
How do you deal with fractions in the combination formula?
 A: The number of combinations is usually found using the factorial function $n!$ There is a standard mathematical function, called the Gamma function, which generalizes the factorial function to all real numbers (in fact all complex numbers) except for negative integers. For positive integers, $\Gamma(n)=(n-1)!$. Note $\Gamma(1)=0!=1$.
The gamma function has the properties that $\Gamma(1)=1$ and that $\Gamma(x+1)=x\Gamma(x)$.
With the usual definition of $~\displaystyle \binom{n}{r}$ as $~\displaystyle  \frac{n!}{(n-r)!\space r!}$ , two of the factorials will involve the gamma function because $n$ and $n-r$ are fractions, while the third won't need to because $r$ is an integer.
Now we will use $\Gamma(1\frac{1}{3})$ instead of $(\frac{1}{3})!$ and we will use $\Gamma(-\frac{2}{3})$ instead of $(-1\frac{2}{3})!$ .
Because of the second property of the Gamma function, $$\Gamma(1\frac{1}{3})=\frac{1}{3}\times\Gamma(\frac{1}{3})=\frac{1}{3}\times-\frac{2}{3}\times\Gamma(-\frac{2}{3})$$ and so $~\displaystyle \binom{n}{r}=\frac{n!}{(n-r)!\space r!}=\frac{\Gamma(1\frac{1}{3})}{\Gamma(-\frac{2}{3})\space 2!}=\frac{\frac{1}{3}\times-\frac{2}{3}\times\Gamma(-\frac{2}{3})}{\Gamma(-\frac{2}{3})\space 2!}=\frac{\frac{1}{3}\times-\frac{2}{3}}{1\times2}=-\frac{1}{9}$
which is the given answer.
A: A generalisation of the binomial coefficient is given for real (or even complex) values $\alpha$ by
\begin{align*}
\binom{\alpha}{n}:=
\begin{cases}
\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}&n\geq 0\tag{1}\\
0&n<0
\end{cases}
\end{align*}
Using (1) we can write
\begin{align*}
\color{blue}{\binom{\frac{1}{3}}{2}}
=\frac{\frac{1}{3}\cdot\left(-\frac{2}{3}\right)}{2\cdot 1}
&\color{blue}{=-\frac{1}{9}}
\end{align*}
The formula (1) can be found for instance as formula (5.1) in Concrete Mathematics by R. L. Graham, D. Knuth and O. Patashnik.
A: 
How do you calculate combinations with fractions?

By generalizing the normal definition of $~\displaystyle \binom{n}{k}$.
When $n \in \Bbb{Z^+}$ and $k \in \{0,1,2,\cdots, n\}$,
then you have that
$$\binom{n}{k} = \frac{n!}{k![(n-k)!]}. \tag1 $$
Re-express (1) above as
$$\frac{n!}{(n-k)!} \times \frac{1}{k!}. \tag2 $$
Then, for $k > 0$, re-express (2) above as
$$n \times (n-1) \times (n-2) \times \cdots \times (n+1-k) \times \frac{1}{k!}. \tag3 $$
The re-expression in (3) above provides the necessary definition of $~\displaystyle \binom{1/3}{2}$.  Note also, that when $n$ is a positive integer, that the definition in (3) remains consistent with the standard definition of $~\displaystyle \binom{n}{k}.$
