Solving equations which includes factorials I tried solve this question:
If $x$ things can be arranged in $m$ ways, $x-2$ things can be arranged in $n$ ways and $x-6$ things can be arranged in $p$ ways and $m = 30np$, then find $x$.
$m = 30np$
$\therefore x! = 30(x-2)!(x-6)!$
$\therefore x(x-1)(x-2)! = 30(x-2)!(x-6)!$
$\therefore x(x-1) = 30(x-6)!$
In the last equation, if you use try and error method, then you will find $6$ is answer for $x$...
But is there a way to find $x$ mathematically?

$x(x-1) = 30(x-6)!$ Suppose $x > 6$
$\frac {x(x+1)}{x-6} = \frac {x^2 - 6x + 7x-42 + 42}{x-6} = x + 7 + \frac {42}{x-6} = 30(x-7)!$  Suppose $x > 7$.
$\frac {x(x+1)}{(x-6)(x-7)} = 1 + \frac {14}{x-7} + \frac{42}{(x-6)(x-7)} = 30(x-8)!$
But $1 + \frac {14}{x-7} + \frac{42}{(x-6)(x-7)}\le 1 + 14 + 21 = 36$ so $(x-8)! = 1$.
Or in other words $x \le 9$.
But $x(x+1) = 30(x-6)!$ so $5|x(x+1)$ so $5|x$ or $5|x-1$ so $x = 6$ or $x=5$. $x$ can't be $5$.
or
Basically $x(x-1) \approx x^2$ and $(x-6)! \approx (\frac x2)^{x-6}$ should indicate $x-6 \approx 2$.  Knowing that $30|x(x-1)$ pinpoints it to $x = 6$.
