Question

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

• A. $p$ divides $N$
• B. $p^2$ divides $N$
• C. $p$ cannot divide $N$
• D. none of these

Answer 1

Answer: (A)

$N={}^{2n}{C}_a=\frac{(2n)!}{(n!{)}^2}=\frac{(n+1)(n+2)...(n+n)}{(n!)}$
$\Rightarrow \left(n!\right)N=\left(n+1\right)\left(n+2\right)...\left(n+n\right)$
Since m$<p<2n$. so p divides $(n+1)(n+2)...(n+n)$