Prove that $q$ divides (${}_q \mathrm{C}_x$) where $q$ is an odd prime and $x \in \Bbb{Z}$ is such that $1\lt x \lt q$.

Question

I believe I need to show that $\frac{q!}{x!(q-x)!} = aq $ where $a\in \mathbb{Z}$ How can I show this?

Answer 1

There is the identity ${n \choose k } = {n \choose n-k}$, which means we may assume without loss of generality that $q - x \leq x$. Then clearly the fraction is $\dfrac{q(q-1)\cdots (q-x+1)}{(q-x)(q-x-1) \cdots (1)}$.

Since $q$ is a prime number, you cannot write it as a multiple of two proper factors greater than $1$. Notice that since $q-x \lt q$ (because by assumption $x \gt 1$), the only way to "divide out" the factor of $q$ in the numerator would be to assemble (compositely) $q$ from a number of smaller factors, but this can't be done by our first sentence in this paragraph.

$\square$