Question

1)  $C^{2} _{x-1} = 6$

2) $A^{3} _{x+2} = 6x(x+1)$

3)$frac{A_{x} ^{4} * P_{x-4}}{ P_{x-2} }$ = 42

Answer 1

$1);C_{x-1}^2=6\frac{(x-1)!}{2!(x-1-2)!}=6\frac{(x-1)!}{2!(x-3)!}=6\frac{(x-2)(x-1)}{2!}=6\x^2-3x+2=12\x^2-3x-10=0\D=9+4cdot10=49\x_1=-2,;x_2=5\2);A_{x+2}^2=6x(x+1)\frac{(x+2)!}{(x+2-2)!}=6x(x+1)\frac{(x+2)!}{x!}=6x(x+1)\frac{(x+1)(x+2)}{x}=6x(x+1)\frac{(x+2)}{x}=6x\x+2=6x^2\6x^2-x-2=0\D=1+4cdot6cdot2=49\x_1=-frac12,;x_2=frac23$
$3.;frac{A_x^4cdot P_{x-4}}{P_{x-2}}=42\frac{frac{x!}{(x-4)!}cdot(x-4)!}{(x-2)^!}=42\frac{x}{(x-2)!}=42\(x-1)x=42\x^2-x-42=0\D=1+4cdot42=169\x_1=-6,;x_2=7$