# How to solve 0.5 choose 4?

I was solving this problem for homework. It says, in the problem, that if n is positive you use the generalized definition of binomial coefficients. In my case, n is positive so I just plugged n= 0.5 and r=4 into the equation n!/r!(n-r)!. However, now I'm having issues solving (n-r)! because I'm having to take the factorial of a negative number. Can someone just explain how I would go about solving this part?

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– Rory Daulton Aug 31 '14 at 19:42

$${n \choose k}=\frac{n(n-1)\cdots(n-k+1)}{k!}$$

You can apply this definition to noninteger $n$, so here

$${\frac12 \choose 4}=\frac{\frac12(\frac12-1)(\frac12-2)(\frac12-3)}{24}=\frac{\frac12(-\frac12)(-\frac32)(-\frac52)}{2^3\cdot3}=-\frac5{128}$$

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They had the same exact definition in the book, but I plugged it in wrong. Thank you! – Ayoshna Aug 31 '14 at 19:43

I believe it goes something like this:

$\dbinom{.5}{4}=\dfrac{(.5)!}{4!(.5-4)!}=\dfrac{(.5)(-.5)(-1.5)(-2.5)(-3.5)!}{4!(-3.5)!}=\\ \quad \dfrac{(.5)(-.5)(-1.5)(-2.5)}{4!}=-\dfrac{15/16}{24}=-\dfrac{5}{128}$

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