Solve the equation: $\;$ $C^{x}_3 + C^{x}_{4} = 11 \times C^{x+1}_{2}$, $\;\;$ $x \in N$
$C^{x}_3 + C^{x}_{4} = 11 \times C^{x+1}_{2}$
i.e. $\;$ $C^{x+1}_{4} = 11 \times C^{x+1}_{2}$ $\;\;\;$ $\left[\because \; C^{n}_{m} + C^{n}_{m + 1} = C^{n + 1}_{m + 1}\right]$
i.e. $\;$ $\dfrac{\left(x + 1\right)!}{4! \left(x + 1 - 4\right)!} = 11 \times \dfrac{\left(x + 1\right)!}{2! \left(x + 1 - 2\right)!}$
i.e. $\;$ $2! \left(x - 1\right)! = 11 \times 4! \left(x - 3\right)!$
i.e. $\;$ $2! \left(x - 1\right) \left(x - 2\right) \left(x - 3\right)! = 11 \times 4 \times 3 \times 2! \left(x - 3\right)!$
i.e. $\;$ $\left(x - 1\right) \left(x - 2\right) = 12 \times 11$
$\implies$ $x - 1 = 12$ $\;$ or $\;$ $x - 2 = 11$
$\implies$ $x = 13$