Решите пятый номер уравнений

Решите задачу:

$A_x^4= \frac{x!}{(x-4)!} = (x-3)(x-2)(x-1)x \\ P_{x-2}=(x-2)!\\ P_{x-4}=(x-4)!\\x\ \textgreater \ 4\\(x-3)(x-2)(x-1)x*(x-4)! \leq 42*(x-2)!\\ x*(x-1)(x-2)(x-3) \leq42(x-2)(x-3) \\ x(x-1) \leq 42=6*7=\ \textgreater \ x \leq 7\\ x\in (4;7]$

$\frac{(n+1)!}{(n-1)!} +n=24\\n\ \textgreater \ 0\\ n(n+1)+n=24\\ n^2+2n-24=0\\ D=4+96=100\\ n=(-2+ 10)/2=4$