Solve the equation: $\;$ $C^{x+1}_{x-4} = \dfrac{7}{15} P^{x+1}_{3}$, $\;$ $x \in N$
$C^{x+1}_{x-4} = \dfrac{7}{15} P^{x+1}_{3}$
i.e. $\;$ $\dfrac{\left(x + 1\right)!}{\left(x - 4\right)! \left(x + 1 - x + 4\right)!} = \dfrac{7}{15} \times \dfrac{\left(x + 1\right)!}{\left(x + 1 - 3\right)!}$
i.e. $\;$ $\dfrac{15}{\left(x - 4\right)! \times 5!} = \dfrac{7}{\left(x - 2\right)!}$
i.e. $\;$ $\dfrac{15}{\left(x - 4\right)! \times 5 \times 4 \times 3 \times 2} = \dfrac{7}{\left(x - 2\right) \left(x - 3\right) \left(x - 4\right)!}$
i.e. $\;$ $\dfrac{1}{8} = \dfrac{7}{\left(x - 2\right) \left(x - 3\right)}$
i.e. $\;$ $\left(x - 2\right) \left(x - 3\right) = 8 \times 7$
$\implies$ $x - 2 = 8$ $\;$ or $\;$ $x - 3 = 7$
$\implies$ $x = 10$