Solve the equation: $\;$ $P^{x + 1}_{3} + C^{x + 1}_{x - 1} = 14 \left(x + 1\right)$, $\;$ $x \in N$
$P^{x + 1}_{3} + C^{x + 1}_{x - 1} = 14 \left(x + 1\right)$
i.e. $\;$ $\dfrac{\left(x + 1\right)!}{\left(x + 1 - 3\right)!} + \dfrac{\left(x + 1\right)!}{\left(x - 1\right)! \left(x + 1 - x + 1\right)!} = 14 \left(x + 1\right)$
i.e. $\;$ $\dfrac{\left(x + 1\right)!}{\left(x - 2\right)!} + \dfrac{\left(x + 1\right)!}{\left(x - 1\right)! 2!} = 14 \left(x + 1\right)$
i.e. $\;$ $\dfrac{\left(x + 1\right) x \left(x - 1\right) \left(x - 2\right)!}{\left(x - 2\right)!} + \dfrac{\left(x + 1\right) x \left(x - 1\right)!}{2 \left(x - 1\right)!} = 14\left(x + 1\right)$
i.e. $\;$ $x \left(x + 1\right) \left(x - 1\right) + \dfrac{x \left(x + 1\right)}{2} = 14\left(x + 1\right)$
i.e. $\;$ $x \left(x - 1 + \dfrac{1}{2}\right) = 14$
i.e. $\;$ $\dfrac{x \left(2x - 1\right)}{2} = 14$
i.e. $\;$ $2x^2 - x - 28 = 0$
i.e. $\;$ $2 \left(x - 4\right) \left(x + 3.5\right) = 0$
$\implies$ $x = 4$, $\;$ or $\;$ $x = -3.5$
$\because \;$ $x \in N$ $\;\;\;$ [given]
$\implies$ $x = 4$ $\;$ is the required solution.