Prove that If $p$ is prime s.t. $0<n\leq p$ , then $p|[{p! \over {(p-n)!(n)!}}]$.

Question

Prove that If $p$ is prime s.t. $0<n\leq p$ , then $p|[{p! \over {(p-n)!(n)!}}]$. I know that if $p|q$ , then $q=kp$, for some integer number $k$. But I don’t know how to prove that $p$ divided like above. Is it working to use proof by induction?

Answer 1

Hint (assuming you have $n<p$): Show that $p$ does divide the numerator, but doesn't divide either of the factors in the denominator.

Answer 2

$(p-n)!$ and $n!$ do not contain $p$ as a factor, so neither of them can divide $p$. However $\binom{p}{n}$ is an integer, containing $p$ as a factor. So $p$ divides $\binom{p}{n}$.

Answer 3

• $p\mid p!$, since $p!=p\cdot (p-1)!$
• If $p\mid (p-n)!$, then $p$ divides atleast one of the factors $p-n, p-(n+1), 3, 2, 1$. But this is impossible, since they are all less than $p$. Same argument with $n!$. Since $p$ is a prime number, $p\not\mid (p-n)!n!$.