I was wondering if it's possible to rearrange for $y$ or $x$ in the following equation containing factorials?
$$x(y!)(x!)y = q $$
I try to solve for $y$ as
$$y \cdot y!= \frac{q}{x \cdot x!}$$
Playing around I only complicate it:
$$y^2(y-1)!=\frac{q}{x \cdot x!}$$
$$y^{2} \Gamma\left(y\right)=\frac{q}{x \cdot x!}$$
I alternatively try to guess if there's a way to simplify $y \cdot y!$ and come to a close yet incorrect guess for the LHS:
$y \cdot y!$ ~ $(y+1)!-1$
I know it's possible to take the inverse factorial as outlined here: Inverse of a factorial, but in order to reach this step I have to have only one factorial variable on the LHS.
I'm ultimately trying to isolate y or x on LHS. Any hints welcome.
Considering that we are talking about 'factorial' and not any 'gamma function'.
Factorial is only defined for positive integer values. So, the variables given here (x,y) must be positive integers as well.
Now, you would realise that the solution should be just $x=1$ and $y=1$ as any other integer value would overshoot the value $1$.