Solve the equation: $\;$ $C^{x + 1}_{3} : C^{x}_{4} = 6 : 5$, $\;$ $x \in N$
$C^{x + 1}_{3} : C^{x}_{4} = 6 : 5$
i.e. $\;$ $\dfrac{\left(x + 1\right)!}{3! \left(x + 1 - 3\right)!} : \dfrac{x!}{4! \left(x - 4\right)!} = 6: 5$
i.e. $\;$ $\dfrac{\left(x + 1\right)!}{3! \left(x - 2\right)!} \times \dfrac{4! \left(x - 4\right)!}{x!} = \dfrac{6}{5}$
i.e. $\;$ $\dfrac{\left(x + 1\right) x!}{3! \left(x - 2\right) \left(x - 3\right) \left(x - 4\right)!} \times \dfrac{4 \times 3! \left(x - 4\right)!}{x!} = \dfrac{6}{5}$
i.e. $\;$ $\dfrac{4 \left(x + 1\right)}{\left(x - 2\right) \left(x - 3\right)}= \dfrac{6}{5}$
i.e. $\;$ $\dfrac{2 \left(x + 1\right)}{x^2 - 5x + 6} = \dfrac{3}{5}$
i.e. $\;$ $10x + 10 = 3x^2 - 15x + 18$
i.e. $\;$ $3x^2 - 25x + 8 = 0$
i.e. $\;$ $\left(x - 8\right) \left(3x - 1\right) = 0$
i.e. $\;$ $x = 8$ $\;$ or $\;$ $x = \dfrac{1}{3}$
$\because \;$ $x \in N$ $\;\;\;$ [given]
$\implies$ $x = 8$ $\;$ is the required solution.