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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 $$-4 \choose 3$$

I am definitely not understanding something here, please help.

mechanodroid
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Tagwoh
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3 Answers3

34

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:

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

Brian M. Scott
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15

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

Will Orrick
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    That explanation is freaking awesome. One up! – Shocky2 Jul 23 '17 at 09:12
  • 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? – Kurapika Feb 03 '24 at 22:15
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    @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, $0$. – Will Orrick Feb 03 '24 at 22:30
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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.

Geoff Robinson
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