# Any way to simplify $\binom{n-x}{k} / \binom{n}{k}\,$?

I tried Wolfram but it just gave me the same thing. I feel like there should be a way to process this. Any thoughts?

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That'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. –  EuYu Oct 17 '12 at 14:56

$$\binom{n-x}{k} \Big/ \binom{n}{k} = \frac{(n-x)!}{(n-x-k)!k!} \frac{(n-k)!k!} {n!} = \frac{(n-x)^\underline{k}}{n^\underline{k}}$$ 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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The second simplification also assumes $x$ is integer, while its name suggests it might not be. You first line also seems to depend on $x$ integer, but it doesn't because $\binom xk=\frac{x^{\underline k}}{k!}$ for any $x$, which incidentally shortens the derivation. –  Marc van Leeuwen Oct 23 '12 at 13:09
The falling factorial has more standard notations, and it is called the "Pochhammer symbol", usually denoted $(x)_n$. en.wikipedia.org/wiki/Pochhammer_symbol –  Patrick Da Silva Nov 24 '12 at 1:33
@Pat, it's the rising factorial that is equivalent to the Pochhammer symbol; the falling factorial is related, but different. –  Guess who it is. Nov 24 '12 at 14:04

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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Meta-comment: Adding signatures to answers is generally frowned upon. (From the FAQ: "Please don’t use signatures or taglines in your posts, or they will be removed.") –  Douglas S. Stones Oct 23 '12 at 12:52
@DouglasS.Stones: and so it shall be... –  robjohn Nov 24 '12 at 1:29