Solve the inequation: $\;$ $C^{x-1}_{4} - C^{x-1}_{3} - \dfrac{5}{4} P^{x-2}_{2} < 0$, $\;$ $x \in N$
$C^{x-1}_{4} - C^{x-1}_{3} - \dfrac{5}{4} P^{x-2}_{2} < 0$
i.e. $\;$ $C^{x-1}_{4} - C^{x-1}_{3} < \dfrac{5}{4} P^{x-2}_{2}$
i.e. $\;$ $\dfrac{\left(x - 1\right)!}{4! \left(x - 1 - 4\right)!} - \dfrac{\left(x - 1\right)!}{3! \left(x - 1 - 3\right)!} < \dfrac{5 \times \left(x - 2\right)!}{4 \times \left(x - 2 - 2\right)!}$
i.e. $\;$ $\dfrac{\left(x - 1\right)!}{4! \left(x - 5\right)!} - \dfrac{\left(x - 1\right)!}{3! \left(x - 4\right)!} < \dfrac{5 \times \left(x - 2\right)!}{4 \times \left(x - 4\right)!}$
i.e. $\;$ $\dfrac{\left(x - 1\right) \left(x - 2\right)!}{24 \left(x - 5\right)!} - \dfrac{\left(x - 1\right) \left(x - 2\right)!}{6 \left(x - 4\right) \left(x - 5\right)!} < \dfrac{5 \times \left(x - 2\right)!}{4 \left(x - 4\right) \left(x - 5\right)!}$
i.e. $\;$ $\dfrac{x-1}{12} - \dfrac{x-1}{3 \left(x - 4\right)} < \dfrac{5}{2 \left(x - 4\right)}$
i.e. $\;$ $\dfrac{\left(x - 1\right) \left(x - 4\right) - 4 \left(x - 1\right)}{12 \left(x - 4\right)} < \dfrac{5}{2 \left(x - 4\right)}$
i.e. $\;$ $\dfrac{x^2 - 5x + 4 - 4x + 4}{6} < 5$
i.e. $\;$ $x^2 - 9x + 8 < 30$
i.e. $\;$ $x^2 - 9x - 22 < 0$
i.e. $\;$ $\left(x - 11\right) \left(x + 2\right) < 0$
$\implies$ $x - 11 <0$ $\;$ and $\;$ $x + 2 > 0$ $\; \;$ OR $\;\;$ $x - 11 > 0$ $\;$ and $\;$ $x + 2 < 0$
i.e. $\;$ $x < 11$ $\;$ and $\;$ $x > -2$ $\;\;$ OR $\;\;$ $x > 11$ $\;$ and $\;$ $x < -2$
$\because \;$ $x \in N$ $\implies$ $x = \left\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\right\}$
Now, as per the question, $x - 1$ elements are taken $4$ at a time
$\implies$ $x = \left\{5, 6, 7, 8, 9,10\right\}$