Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $5 C^{n}_{3} < C^{n+2}_{4}$, $\;$ $n \in N$

$5 C^{n}_{3} < C^{n+2}_{4}$

i.e. $\;$ $\dfrac{5 \times n!}{3! \left(n - 3\right)!} < \dfrac{\left(n + 2\right)!}{4! \left(n + 2 - 4\right)!}$

i.e. $\;$ $\dfrac{5 \times n!}{3! \left(n - 3\right)!} < \dfrac{\left(n + 2\right)!}{4! \left(n - 2\right)!}$

i.e. $\;$ $\dfrac{5 \times n!}{3! \left(n - 3\right)!} < \dfrac{\left(n + 2\right) \left(n + 1\right) n!}{4 \times 3! \left(n - 2\right) \left(n - 3\right)!}$

i.e. $\;$ $5 < \dfrac{\left(n + 2\right) \left(n + 1\right)}{4 \left(n - 2\right)}$

i.e. $\;$ $20n - 40 < n^2 + 3n + 2$

i.e. $\;$ $0 < n^2 - 17n + 42$

i.e. $\;$ $n^2 - 17n + 42 > 0$

i.e. $\;$ $\left(x - 14\right) \left(x - 3\right) > 0$

i.e. $\;$ $x - 14 > 0$ $\;$ or $\;$ $x - 3 > 0$

i.e. $\;$ $x > 14$ $\;$ or $\;$ $x > 3$

$\implies$ $x > 14, \; n \in N$