Комбинаторика ... не могу решить.... помогите

$1) x \geq 2\\C_{x-2}^{x-4}= \frac{(x-2)!}{(x-4)!*2!} =(x-3)(x-2)/2\\ (x-3)(x-2)=4x^2-14x-6\\ x^2-5x+6 = 4x^2-14x-6\\ 3x^2-9x-12=0\\ x^2-3x-4=0\\ D=9+16=25\\ x = (3+5)/2=4$

$2) x \geq 2\\2C_{x-1}^2= \frac{2(x-1)!}{(x-3)!2!} =(x-2)(x-1)\\ 2C_x^{x-2} = \frac{2x!}{(x-2)!2!} =(x-1)x\\ (x-1)(x-2) + (x-1)x = (x-1)(x+1)\\ (x-1)(x-2+x-x-1)=0\\ x_1 = 1\\ x-3=0\\ x_2 =3$

Ответ: x = 3

$3)x \geq 0;y \geq 0\\ C_x^y= \frac{x!}{y!(x-y)!} = C_x^{y+2}= \frac{x!}{(y+2)!(x-y-2)!} \\ y!(x-y)!=(y+2)!(x-y-2)!\\ (x-y-1)(x-y) = (y+1)(y+2)\\ \\ C_x^2= \frac{(x-1)x}{2} =66\\ x(x-1) = 132\\ x^2 - x - 132 = 0\\ D = 1 + 528 = 529 = 23^2\\ x = (1+23)/2 = 12\\ \\ (12-y-1)(12-y)=(y+1)(y+2)\\ 132-23y+y^2=y^2+3y+2\\ 26y=130\\ y=5$
Ответ: (12;5)