TITLE: The divisibility of a particular case of $\binom{x}{y}$ (when $(x,y) = 1$)
QUESTION [5 upvotes]: Suppose $x + 2y = k$ where $k$ is prime and $x > y > 0$.
Prove that $x|\binom{x}{y}$. That is, prove that $\binom{x}{y}$ is divisible by $x$.
The only leads I have are that, since $x$ and $y$ must be relatively prime -- or else $k$ would not be prime -- in the expansion of the binomial coefficient:
$$\frac{(x)(x-1)\dots(x-y+1)}{(y)(y-1)\dots 1}$$, there is necessarily some value in $\{x-1 \dots x-y+1\}$ that is divisible by $y$. This is true, because given any consecutive $y$ numbers, at least one must be divisible $y$, and we know that $x$ is not that number.
Edit: Apparently this is answered here When is $\binom{n}{k}$ divisible by $n$?
I'm not seeing how the result is obvious, however, for the particular case that $(n,k) = 1$.

REPLY [2 votes]: We can use this nice identity
$$
\binom{x}{y}\binom{x-y}{h} = \binom{x}{h}\binom{x-h}{y} 
$$
to obtain
$$
\binom{x}{y}y = \binom{x}{y}\binom{y}{1} = \binom{x}{1}\binom{x-1}{x-y} = x\binom{x-1}{x-y}.
$$
Now we see that $x$ divides right side, so it must divide also left side. But since $(x,y)=1$, $x$ must divide $\binom{x}{y}$.