# Binomial coefficients $1/2\choose k$

I don't understand questions that involve a binomial expression where you have a fraction choose $k$ or a negative number choose $k$. I understand and am able to do it when there are no fractions and they are all positive. We learned the generalized formula but I get the wrong answer when the question involves fractions or negative numbers.

eg: $$2/3 \choose 2$$ or $$-4 \choose 3$$

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, $\binom{1/2}{3}=\frac{(1/2)(1/2-1)(1/2-2)}{3\cdot2\cdot1}=\frac{(1/2)(-1/2)(-3/2)}{3\cdot2\cdot1}=\frac{1}{16}.$

• That explanation is freaking awesome. One up! Commented Jul 23, 2017 at 9:12
• If you wanted to compute something like 1.5 choose 0, then what would the falling factorial look like? Will we have only one term in the denominator or is this calculation not possible? Commented Feb 3 at 22:15
• @Kurapika $\binom{x}{0}$, for any $x$, is the empty product divided by the empty product. The empty product is, by definition, $1$, in analogy to how the empty sum is, by definition, $0$. Commented Feb 3 at 22:30

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.