If ${^{\left(n + 2\right)}}{C}_{8} : {^{\left(n - 2\right)}}{P}_{4} = 57 : 16$, find $n$.
Given: $\;$ ${^{\left(n + 2\right)}}{C}_{8} : {^{\left(n - 2\right)}}{P}_{4} = 57 : 16$
i.e. $\;$ $\dfrac{\left(n + 2\right)!}{8! \times \left(n - 6\right)!} \times \dfrac{\left(n - 6\right)!}{\left(n - 2\right)!} = \dfrac{57}{16}$
i.e. $\;$ $\dfrac{\left(n + 2\right) \times \left(n + 1\right) \times \left(n\right) \times \left(n - 1\right)}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2} = \dfrac{57}{16}$
i.e. $\;$ $\left(n + 2\right) \times \left(n + 1\right) \times \left(n\right) \times \left(n - 1\right) = 57 \times 7 \times 6 \times 5 \times 4 \times 3$
i.e. $\;$ $\left(n + 2\right) \times \left(n + 1\right) \times \left(n\right) \times \left(n - 1\right) = 19 \times 3 \times 7 \times 6 \times 5 \times 4 \times 3$
i.e. $\;$ $\left(n + 2\right) \times \left(n + 1\right) \times \left(n\right) \times \left(n - 1\right) = 21 \times 20 \times 19 \times 18$
$\implies$ $n = 19$