Prove this fraction will increase as x and y decrease

Question

Is there a way to prove this fraction will increase as x and y decrease: $$0<x<1$$ $$y>1 $$

$$\frac{xy}{x^2(\frac{y!}{2!(y-2)!})}$$ $\frac{y!}{2!(y-2)!}$ is the combination formula, C(y,2)

Answer 1

$$\frac{a}{(\frac bc)}=\frac{ac}{b}$$ So your expression is: $$\frac{2xy(y-2)!}{x^2y!}=\frac{2y(y-2)!}{xy!}=\frac{2(y-2)!}{x(y-1)!}=\frac{2}{x(y-1)}$$

This clearly doesn't increase as $x,y$ increase, so I believe the question is wrong.

Answer 2

Note that the combination $C(y,2)$ has $(y-2)!$ in the denominator.

On the other hand $$\frac{xy}{x^2(\frac{y!}{2!(y-2)!})} = \frac {2}{x(y-1)}$$ is a decreasing function on the given domain.