Algebra - Elements of Combinatorics

Solve the equation: $\;$ $\dfrac{P_{x + 3}}{P^{x}_{5} \times P_{x-5}} = 720$, $\;\;$ $x \in N$


$\dfrac{P_{x + 3}}{P^{x}_{5} \times P_{x-5}} = 720$

i.e. $\;$ $\dfrac{\left(x + 3\right)!}{\dfrac{x!}{\left(x - 5\right)!} \times \left(x - 5\right)!} = 720$

i.e. $\;$ $\dfrac{\left(x + 3\right)!}{x!} = 720$

i.e. $\;$ $\dfrac{\left(x + 3\right) \left(x + 2\right) \left(x + 1\right) x!}{x!} = 720$

i.e. $\;$ $\left(x + 3\right) \left(x + 2\right) \left(x + 1\right) = 10 \times 9 \times 8$

$\implies$ $x + 3 = 10$ $\;\;$ or $\;\;$ $x + 2 = 9$ $\;\;$ or $x + 1 = 8$

$\implies$ $x = 7$