Solve the equation: $\;$ $3 C^{x+1}_{2} + P_2 \cdot x = 4 P^x_2$, $\;$ $x \in N$
$3 C^{x+1}_{2} + P_2 \cdot x = 4 P^x_2$
i.e. $\;$ $\dfrac{3 \times\left(x + 1\right)!}{2! \left(x + 1 -2\right)!} + 2! \times x = \dfrac{4 \times x!}{\left(x - 2\right)!}$
i.e. $\;$ $\dfrac{3 \times \left(x + 1\right)!}{2 \times \left(x - 1\right)!} + 2x = \dfrac{4 x \left(x - 1\right) \left(x - 2\right)!}{\left(x - 2\right)!}$
i.e. $\;$ $\dfrac{3 \left(x + 1\right) x \left(x - 1\right)!}{2 \left(x - 1\right)!} + 2x = 4 x \left(x - 1\right)$
i.e. $\;$ $\dfrac{3 x \left(x + 1\right)}{2} + 2x = 4x \left(x - 1\right)$
i.e. $\;$ $3x + 3 + 4 = 8x - 8$
i.e. $\;$ $5x = 15$
$\implies$ $x = 3$