Question

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

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

Answer 1

The correct option is A $p$ divides $N$ completely

Solution:

$N={}^{2n}{C}_n=\frac{2n!}{n!n!}=\frac{(n+1)(n+2)(n+3)........\left(2n\right)}{\left(1\right).\left(2\right).......\left(n\right)}$

As $p>n$ and $p<2n,p$ occurs exactly once in the numerator but doesn't occur in the denominator.

It means $p$ divides $N$ completely.

Hence, A is the correct option.