Binomial coefficients $1/2\choose k$

You know that $$\binom{x}k=\frac{x^{\underline k}}{k!}\;,$$ where $x^{\underline k}$ is the falling factorial: $x^{\underline k}=x(x-1)(x-2)\dots(x-k+1)$. Thus,

$$\binom{2/3}2=\frac{(2/3)^{\underline 2}}{2!}=\frac{\left(\frac23\right)\left(\frac23-1\right)}2=\frac{\left(\frac23\right)\left(-\frac13\right)}2=-\frac19\;,$$

and

$$\binom{-4}3=\frac{(-4)^{\underline 3}}{3!}=\frac{(-4)(-4-1)(-4-2)}6=-\frac{4\cdot5\cdot6}6=-20\;.$$

With specific small numbers you can always just do the arithmetic, as I’ve done here. Some more general calculations are also possible without too much difficulty. For instance:

$$\begin{align*} \binom{1/2}n&=\frac{(1/2)^{\underline n}}{n!}\\ &=\frac{\left(\frac12\right)\left(-\frac12\right)\left(-\frac32\right)\dots\left(-\frac{2n-3}2\right)}{n!}\\ &=(-1)^{n-1}\frac{(2n-3)!!}{2^nn!}\\ &=(-1)^{n-1}\frac{2^{n-1}(n-1)!(2n-3)!!}{2^{2n-1}n!(n-1)!}\\ &=(-1)^{n-1}\frac{(2n-2)!!(2n-3)!!}{2^{2n-1}n!(n-1)!}\\ &=\frac{(-1)^{n-1}}{2^{2n-1}n}\frac{(2n-2)!}{(n-1)!^2}\\ &=\frac{(-1)^{n-1}}{2^{2n-1}n}\binom{2n-2}{n-1} \end{align*}$$

$\binom{m}{k}$ is the ratio of two products, both of which contain $k$ factors, and in both of which the factors descend in steps of 1. For example, $\binom{1/2}{3}=\frac{(1/2)(1/2-1)(1/2-2)}{3\cdot2\cdot1}=\frac{(1/2)(-1/2)(-3/2)}{3\cdot2\cdot1}=\frac{1}{16}.$

In the definition $\left( \begin{array}{cc} m \\ k \end{array}\right) = \frac{m(m-1) \ldots (m-k+1)}{1.2.\ldots k }$ it is not necessary that $m$ should be a positive integer, though $k$ is usually taken to be a positive integer, and this formula allows routine evaluation of other types of binomial coefficients.