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I wondered why the expression $$ \binom {2n}{n} $$ is always divisible by each prime p with $$ n < p \leq 2n $$

Now I wanted to proof this but I don't know where I should start...

I can rewrite the expression to: $$ \binom {2n}{n} = \frac{(2n!)}{n!\cdot(2n-n)!}=\frac{(2n)!}{n!^2}$$ but I dont think, that this will really help, so i hope that someone can help me!

Thanks! :)

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    for the number is an integer, and the denominator doesn't contain any prime from $n$ to $2n.$ – 2017-01-02

2 Answers 2

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  • Every prime $n
  • if $p$ is prime, and $n

Then $(2n)!/n!^2$ is multiple of $p$

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    mhh, this was a bit too fast... I have'nt understood it :/ – 2017-01-02
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    could you maybe explain it again, a little more detailed, please? – 2017-01-03
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    @Niklas I just edit – 2017-01-03
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    Thanks, but I dont get the step from the two points to the $$(2n)!/n!^2$$... I think I have blackout but I cant get it... :D – 2017-01-03
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    Suppose $p$ is prime and $n$$(2n)!=1\cdot 2\cdot\cdots\cdot (n-1)\cdot n\cdot\cdots\cdot p\cdot\cdots\cdot (2n-1)\cdot2n.$$ – 2017-01-03
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    This part I understand, but why do I need: $\begin{multline} \\ \end{multline}$ "- if $p$ is prime, and $n$n!$ is not a multiple of $p$ because every divisor $d$ of $n!$ is $d\leq n $d\neq p$" $\begin{multline} \\ \end{multline}$ ?? – 2017-01-03
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    @Laars Helenius, $\begin{multline} \\ \end{multline}$ Could someone maybe explain, why I need the info: $d /=p$? – 2017-01-03
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    @Niklas The idea is: suppose that n! is a multiple of p. but i proved that every divisor d of n! Is

    – 2017-01-03

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    Mhh, okay... I do not really understsand it but I think I simply accept it.. :/ – 2017-01-03
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    Wait, maybe I understand it, if you explain to me, why I have to proof, that n" is not a multiple of p... – 2017-01-03
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Consider any prime $p\le n$. Then $p$ is a factor of both $n!$ and $(2n)!$, say $n!=p\cdot k$ and $(2n)!=p\cdot 2p\cdot \ell$. Then $$ \binom{2n}{n}=\frac{(2n)!}{n!n!}=\frac{p\cdot 2p\cdot \ell}{p^2\cdot k^2}=\frac{2\ell}{k^2} $$ so it is not clear if $p$ divides $\binom{2n}{n}$ when $p\le n$. However, if $n