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 $(n+1)(n+2)...(n+n)$