Question

If $n<p<2n$ and $p$ is prime and $N{=}^{2n}{C}_n$, then

• A. $\frac{p}{N}$
• B. $p$ does not divide $N$
• C. $\frac{p^2}{N}$
• D. $p^2$ does not divide $N$

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{p}{N}$
D $p^2$ does not divide $N$

A,D

$N=\frac{2n!}{n!n!}=\frac{(n+1)(n+2)....\left(2n\right)}{n!}\Rightarrow n!N=(n+1)(n+2)....\left(2n\right)$

$\frac{p}{(n+1)(n+2)....\left(2n\right)}$

as $n\le p\le 2n$

Thus $\frac{p}{n!N}$ or

$\frac{p}{N}$ as n! does not divide p.

If possible, let $\frac{p^2}{N}$

$\Rightarrow \frac{p^2}{n!N}=\frac{p^2}{(n+1)(n+2)....\left(2n\right)}$

or $\frac{p}{(n+1)(n+2)....\left(2n\right)}$

This is not possible as $p^2$ does not divide N.

so A D hold.