Solve the equation: $\;$ $11 C^{x}_{3} = 24 C^{x+1}_{2}$, $\;\;$ $x \in N$
$11 C^{x}_{3} = 24 C^{x+1}_{2}$
i.e. $\;$ $\dfrac{11 \times x!}{3! \left(x - 3\right)!} = \dfrac{24 \times \left(x + 1\right)!}{2! \left(x + 1 - 2\right)!}$
i.e. $\;$ $\dfrac{11 \times x!}{3! \left(x - 3\right)!} = \dfrac{24 \times \left(x + 1\right)!}{2! \left(x - 1\right)!}$
i.e. $\;$ $\dfrac{11 \times x!}{ 3 \times 2! \left(x - 3\right)!} = \dfrac{24 \times \left(x + 1\right) x!}{2! \left(x - 1\right) \left(x - 2\right) \left(x - 3\right)!}$
i.e. $\;$ $\dfrac{11}{3} = \dfrac{24 \left(x + 1\right)}{\left(x - 1\right) \left(x - 2\right)}$
i.e. $\;$ $11 x^2 - 33x + 22 = 72x + 72$
i.e. $\;$ $11 x^2 - 105 x - 50 = 0$
i.e. $\;$ $\left(11x + 5\right) \left(x - 10\right) = 0$
i.e. $\;$ $11x + 5 = 0$ $\;$ or $\;$ $x - 10 = 0$
i.e. $\;$ $x = \dfrac{-5}{11}$ $\;$ or $x = 10$
$\because \;$ $x \in N$ $\implies$ $x = 10$ $\;$ is the required solution.