Solve the equation: $\;$ $P^{x-1}_{2} - C^{x}_{1} = 79$, $\;$ $x \in N$
$P^{x-1}_{2} - C^{x}_{1} = 79$
i.e. $\;$ $\dfrac{\left(x - 1\right)!}{\left(x - 1 - 2\right)!} - \dfrac{x!}{1! \left(x - 1\right)!} = 79$
i.e. $\;$ $\dfrac{\left(x - 1\right)!}{\left(x - 3\right)!} - \dfrac{x!}{1 \times \left(x - 1\right)!} = 79$
i.e. $\;$ $\dfrac{\left(x - 1\right) \left(x - 2\right) \left(x - 3\right)!}{\left(x - 3\right)!} - \dfrac{x \left(x - 1\right)!}{\left(x - 1\right)!} = 79$
i.e. $\;$ $\left(x - 1\right) \left(x - 2\right) - x = 79$
i.e. $\;$ $x^2 - 3x + 2 - x = 79$
i.e. $\;$ $x^2 - 4x -77 = 0$
i.e. $\;$ $x^2 -11x + 7x - 77 = 0$
i.e. $\;$ $x \left(x - 11\right) + 7 \left(x - 11\right) = 0$
i.e. $\;$ $\left(x - 11\right) \left(x + 7\right) = 0$
$\implies$ $x = 11$ $\;$ or $\;$ $x = -7$
$\because \;$ $x \in N$ $\;\;$ [given]
$\implies$ $x = 11$ $\;$ is the required solution.