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I was recently given a question in my homework, and part of the formula asks for 0.5 choose 2, how would I work out the decimal factorial?

0.5 c 2

N. F. Taussig
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Lily V
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1 Answers1

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$$\binom{n}{r}=\frac{n!}{r!(n-r)!}$$ However as $n\notin\Bbb Z$, the typical definition of $n!=1\cdot2\cdots n$ is not applicable. Instead we use the falling factorial $n^{\underline{r}}=n(n-1)\cdots(n-(r-1))$ and the $(n-r)!$ vanishes (To see why, try integer $n$ and $r$ using the falling factorials, and you'll see that $n^{\underline{r}}=\frac{n!}{(n-r)!}$).

So we get: $$\binom{n}{r}=\frac{n^{\underline{r}}}{r!}$$ In the case of $n=\frac 12, r=2$, we get $$\binom{\frac 12}{2}=\frac{(\frac12)(\frac12-1)}{2!}=\frac{-\frac14}{2}=-\frac18$$

Rhys Hughes
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  • 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 22:14
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    Generally $\binom n0 =1$. This should hold, by having no terms in the numerator you just have 1, the identity, and the denominator is $0!=1$. – Rhys Hughes Feb 05 '24 at 00:54