I tried Wolfram but it just gave me the same thing. I feel like there should be a way to process this. Any thoughts?
Any way to simplify $\binom{n-x}{k} / \binom{n}{k}\, $?
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$\begingroup$
combinatorics
binomial-coefficients
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1That's pretty much as simple as it gets. You can cancel a $k!$ if you expand the binomial coefficients out, but I'm not sure if that's _simpler_. – 2012-10-17
2 Answers
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$\binom{n-x}{k} \Big/ \binom{n}{k} = \frac{(n-x)!}{(n-x-k)!k!} \frac{(n-k)!k!} {n!} = \frac{(n-x)^\underline{k}}{n^\underline{k}}$ When $x < k$ the above can be further simplified to $\frac{(n-k)^\underline{x}}{n^\underline{x}}$
Here $x^\underline{y} = x(x-1)\cdots(x-y+1)$ is the falling factorial.
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0@Pat, it's the *rising* factorial that is equivalent to the Pochhammer symbol; the falling factorial is related, but different. – 2012-11-24
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You can use Sterling's formulae to "approximate", if that is deemed as a simplification. Notice that both lower and upper bounds are possible, and those bounds are pretty tight, so, those "might" suit your purpose. Please see at: http://en.wikipedia.org/wiki/Stirling's_approximation
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0@DouglasS.Stones: and so it shall be... – 2012-11-24