Question

Помогите решить задание 2 под буквой б, первая цифра 30

Answer 1

$30\cdot C_{x-3}^{x-9}=19\cdot A_{x-4}^4\\\\\\\star \; \; C_{n}^{m}=\frac{n!}{m!(n-m)!}\; \; ,\; \; A_{n}^{m}=\frac{n!}{(n-m)!}\; \; \star \\\\30\cdot \frac{(x-3)!}{(x-9)!\, \cdot (x-3-x+9)!}=19\cdot \frac{(x-4)!}{(x-4-4)!}\\\\30\cdot \frac{(x-3)!}{(x-9)!\, \cdot 6!}=19\cdot \frac{(x-4)!}{(x-8)!}\\\\\frac{(x-3)!\; (x-8)!}{6!\; (x-9)!\; (x-4)!}=\frac{19}{30}\\\\\star \; \; (x-3)!=(x-4)!\, (x-3)\; ,\; \; (x-8)!=(x-9)!\; (x-8)\; \; \star \\\\\frac{(x-4)!(x-3)\; (x-9)!(x-8)}{6!\; (x-9)!\; (x-4)!}=\frac{19}{30}$

$(x-3)(x-8)=\frac{19\cdot 6!}{30}\\\\x^2-11x+24=\frac{19\cdot \; 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{30}\\\\x^2-11x+24=456\\\\x^2-11x-432=0\\\\D=1849=43^2\; ,\; \; x_{1,2}=\frac{11\pm 43}{2}\\\\x_1=-16\notin N\; ,\; \; x_2=27\\\\Otvet:\; \; x=27\; .$