Solve the equation: $\;$ $12 C^{x}_{1} + C^{x + 4}_{2} = 162$, $\;$ $x \in N$
$12 C^{x}_{1} + C^{x + 4}_{2} = 162$
i.e. $\;$ $\dfrac{12 \times x!}{1! \left(x - 1\right)!} + \dfrac{\left(x + 4\right)!}{2! \left(x + 4 - 2\right)!} = 162$
i.e. $\;$ $\dfrac{12 x \left(x - 1\right)!}{\left(x - 1\right)!} + \dfrac{\left(x + 4\right)!}{2 \left(x + 2\right)!} = 162$
i.e. $\;$ $12 x + \dfrac{\left(x + 4\right) \left(x + 3\right) \left(x + 2\right)!}{2 \left(x + 2\right)!} = 162$
i.e. $\;$ $12x + \dfrac{\left(x + 4\right) \left(x + 3\right)}{2} = 162$
i.e. $\;$ $24x + x^2 + 7x + 12 = 324$
i.e. $\;$ $x^2 + 31x - 312 = 0$
i.e. $\;$ $\left(x + 39\right) \left(x - 8\right) = 0$
$\implies$ $x = -39$ $\;$ or $\;$ $x = 8$
$\because \;$ $x \in N$ $\;\;$ [given]
$\implies$ $x = 8$ $\;$ is the required solution.