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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.

Factorial_123
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    Are $x$ and $y$ supposed to be real numbers? Or complex numbers? Or something else? – Servaes Mar 16 '24 at 00:35
  • x and y are real numbers, i am only interested in the real domain. I am interested in re-arranging the equation so that the equation is expressed either in terms of y or in terms of x. As in equation x(y)^2=a would give y=sqrt(a/x) with y solely on LHS. – Factorial_123 Mar 16 '24 at 01:15
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    You may be interested to know that the quantity involving y which you want to solve for, namely $y \cdot y!,$ is identical to the difference between succesive factorials $(y+1)! - y!.$ The relation between factorials and the Gamma function means you can rephrase this as $\Gamma(y+2)=\Gamma(y+1).$ In this sense your quest is a step more involved than the simple inversion of the Gamma function, since you need to invert the difference between two "adjacent" values of Gamma. – coffeemath Mar 16 '24 at 18:02
  • @coffeemath I'm confused on how Γ(y+2)=Γ(y+1) can be equal? If i type it out in desmos i get different curves: https://www.desmos.com/calculator/thozwhp09q – Factorial_123 Mar 21 '24 at 06:43
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    @Factorial_123 That was a typo--- of course since I was putting $(y+1)!-y!$ into Gamma function form there should have been a minus sign between the two adjacent Gammas. – coffeemath Mar 21 '24 at 10:11
  • @coffeemath I tried two different approaches and using inverse Gamma factorials, but it seems to create a bigger equation for me. I don't think I'm using the right approach to isolate y: https://www.desmos.com/calculator/fljaausagv – Factorial_123 Mar 22 '24 at 05:29

1 Answers1

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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 $1$.

JEEAspirant
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  • Maybe I phrased it incorrectly, I don't want the values of y or x. I want to re-arrange the equation $x(y)!y(x)!=1 to become y=same equation or x=same equation. Move y or x to the LHS. – Factorial_123 Mar 16 '24 at 01:12