Question

Комбинаторика решить 3 задания буду очень благодарна)

Answer 1

1) В слове "журнал" 2 гласные буквы

выбрать гласную букву :

$C_{2} ^{1} =2$

2)

$\dfrac{x}{A_{x}^{3}}=\dfrac{1}{12} \\\\A_{x}^{3}=12x\\\\\dfrac{x!}{(x-3)!}=12x \\\\\frac{(x-3)!\cdot (x-2)\cdot(x-1)\cdot x}{(x-3)!} =12x\\\\(x-2)(x-1)=12\\\\x^{2}-3x+2-12=0\\\\x^{2}-3x-10=0\\\\x_{1}=5\\\\x_{2}=-2$

$3) 20A_{x-2}^{3}=A_{x}^{5}\\\\20\cdot\frac{(x-2)!}{(x-2-3)!}=\frac{x!}{(x-5)!} \\\\20\cdot\frac{(x-2)!}{(x-5)!}=\frac{x!}{(x-5)!}\\\\20\cdot\frac{(x-5)!\cdot(x-4)\cdot(x-3)\cdot(x-2)}{(x-5)!}=\frac{(x-5)!\cdot(x-4)\cdot(x-3)\cdot(x-2)\cdot(x-1)\cdot x}{(x-5)!}\\\\20(x-4)(x-3)(x-2)-(x-4)(x-3)(x-2)(x-1)x=0\\\\(x-4)(x-3)(x-2)\cdot[20-x(x-1)]=0\\\\(x-4)(x-3)(x-2)\cdot(20-x^{2}+x)=0\\\\x-4=0 \ \Rightarrow \ x_{1} =4\\\\x-3=0 \ \Rightarrow \ x_{2} =3$

$x-2=0 \ \Rightarrow \ x_{3} =2\\\\20-x^{2}+x=0\\\\x^{2} -x-20=0\\\\x_{4} =5\\\\x_{5}=-4$