[tex]\displaystyle\bf\\A_{x-4} ^{2} =\frac{(x-4)!}{(x-4-2)!} =\frac{(x-4)!}{(x-6)!} =\frac{(x-6)!\cdot(x-5)\cdot(x-4)}{(x-6)!} =(x-5)(x-4)\\\\\\C_{x-3} ^{2} =\frac{(x-3)!}{2!\cdot(x-3-2)!} =\frac{(x-3)!}{1\cdot 2\cdot (x-5)!} =\frac{(x-5)!\cdot(x-4)\cdot(x-3)}{2\cdot (x-5)!} =\\\\\\=\frac{(x-4)(x-3)}{2} \\\\\\A_{x-4} ^{2} -C_{x-3}^{2} -20\\\\\\(x-5)(x-4)-\frac{(x-4)(x-3)}{2} < 20\\\\\\2(x-5)(x-4)-(x-4)(x-3)-40 < 0\\\\\\2(x^{2} -9x+20)-(x^{2} -7x+12)-40 < 0[/tex]

[tex]\displaystyle\bf\\2x^{2} -18x+40-x^{2} +7x-12-40 < 0\\\\\\x^{2}-11x-12 < 0\\\\\\(x-12)(x+1) < 0[/tex]

+ + + + + ( - 1) - - - - - (12)+ + + + +

                    //////////

[tex]\displaystyle\bf\\x\in\Big(-1 \ ; \ 12\Big)[/tex]

x - должен быть > 0 , поэтому окончательный ответ :

[tex]\displaystyle\bf\\x\in\Big(0 \ ; \ 12\Big)[/tex]