I was just wondering how I would evaluate something like:

$\displaystyle {\frac{1}{2} \choose 1}$ (1/2 choose 1)

or

$\displaystyle {\frac{1}{2} \choose 2}$ (1/2 choose 2)

I've never seen fractions in binomial coefficients.

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- Sep 19th 2010, 12:41 PMeXistEvaluting Binomial Coefficients that contain fractions?
I was just wondering how I would evaluate something like:

$\displaystyle {\frac{1}{2} \choose 1}$ (1/2 choose 1)

or

$\displaystyle {\frac{1}{2} \choose 2}$ (1/2 choose 2)

I've never seen fractions in binomial coefficients. - Sep 20th 2010, 05:29 AMHallsofIvy
Wikipedia

Binomial coefficient - Wikipedia, the free encyclopedia

talks about that way down the page at "Binomial Coefficient with n= 1/2".

Essentially, you write $\displaystyle \begin{pmatrix}n \\ m\end{pmatrix}$ as $\displaystyle \frac{n!}{m!(n-m)!}= \frac{n(n-1)(n- 2)\cdot\cdot\cdot(n-m+1)}{m(m-1)(m-2)\cdot\cdot\cdot(3)(2)(1)}$ with the difference that since n is not an integer, you never end at "1" or "n- m+1". Instead, you get an infinite product which must be shown to have a finite value.

Wikipedia gives, explicitely,

$\displaystyle \begin{pmatrix}\frac{1}{2} \\ n \end{pmatrix}= \begin{pmatrix}2k+1 \\ k\end{pmatrix}\frac{(-1)^{k+1}(k+1)}{2^{2n}(2k-1)(2k+1)}$