Solve the inequation: $\;$ $C^{x+1}_{x-1} < 21$, $\;$ $x \in N$
$C^{x+1}_{x-1} < 21$
i.e. $\;$ $\dfrac{\left(x + 1\right)!}{\left(x - 1\right)! \left(x + 1 - x + 1\right)!} < 21$
i.e. $\;$ $\dfrac{\left(x + 1\right) x \left(x - 1\right)!}{\left(x - 1\right)! 2!} < 21$
i.e. $\;$ $\left(x + 1\right) x < 42$
i.e. $\;$ $x^2 + x - 42 < 0$
i.e. $\;$ $\left(x + 7\right) \left(x - 6\right) < 0$
i.e. $\;$ $x + 7 < 0$ $\;$ and $\;$ $x - 6 > 0$ $\;\;$ OR $\;\;$ $x + 7 > 0$ $\;$ and $\;$ $x - 6 < 0$
i.e. $\;$ $x < -7$ $\;$ and $\;$ $x > 6$ $\;\;$ OR $\;\;$ $x > -7$ $\;$ and $\;$ $x < 6$
$\implies$ $x = \left\{1, 2, 3, 4, 5\right\}$ $\;$ $\because \; x \in N$