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

Question

Математика

Нонн

6 год назад

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

ОТВЕТЫ

Valeriy

Jun 30, 2019

$1) x \geq 2\\C_{x-2}^{x-4}= \frac{(x-2)!}{(x-4)!*2!} =(x-3)(x-2)/2\\amp;#10;(x-3)(x-2)=4x^2-14x-6\\amp;#10;x^2-5x+6 = 4x^2-14x-6\\amp;#10;3x^2-9x-12=0\\amp;#10;x^2-3x-4=0\\amp;#10;D=9+16=25\\amp;#10;x = (3+5)/2=4$

$2) x \geq 2\\2C_{x-1}^2= \frac{2(x-1)!}{(x-3)!2!} =(x-2)(x-1)\\amp;#10;2C_x^{x-2} = \frac{2x!}{(x-2)!2!} =(x-1)x\\amp;#10;(x-1)(x-2) + (x-1)x = (x-1)(x+1)\\amp;#10;(x-1)(x-2+x-x-1)=0\\amp;#10;x_1 = 1\\amp;#10;x-3=0\\amp;#10;x_2 =3$

Ответ: x = 3

$3)x \geq 0;y \geq 0\\amp;#10;C_x^y= \frac{x!}{y!(x-y)!} = C_x^{y+2}= \frac{x!}{(y+2)!(x-y-2)!} \\amp;#10;y!(x-y)!=(y+2)!(x-y-2)!\\amp;#10;(x-y-1)(x-y) = (y+1)(y+2)\\amp;#10;\\amp;#10;C_x^2= \frac{(x-1)x}{2} =66\\amp;#10;x(x-1) = 132\\amp;#10;x^2 - x - 132 = 0\\amp;#10;D = 1 + 528 = 529 = 23^2\\amp;#10;x = (1+23)/2 = 12\\amp;#10;\\amp;#10;(12-y-1)(12-y)=(y+1)(y+2)\\amp;#10;132-23y+y^2=y^2+3y+2\\amp;#10;26y=130\\amp;#10;y=5$
Ответ: (12;5)

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