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Answer 1

$\displaystyle\bf\\C_{n} ^{3} =4C_{n-2} ^{2} \\\\\\\frac{n!}{3!\cdot(n-3)!} =4\cdot\frac{(n-2)!}{2!\cdot(n-2-2)!} \\\\\\\frac{(n-3)!\cdot(n-2)\cdot(n-1)\cdot n}{1\cdot 2\cdot 3\cdot(n-3)!} =4\cdot\frac{(n-4)!\cdot(n-3)\cdot(n-2)}{1 \cdot 2\cdot (n-4)!} \\\\\\\frac{(n-2)\cdot(n-1)\cdot n}{6} -2(n-3)\cdot (n-2)=0\\\\\\(n-2)(n-1)n-12(n-3)(n-2)=0\\\\\\\Big(n-2\Big)\cdot\Big[\Big(n-1)\cdot n-12\Big(n-3\Big)\Big]=0\\\\\\n-2=0 \ \ \ \Rightarrow \ \ \ n=2$

$\displaystyle\bf\\n^{2}-n-12n+36=0\\\\\\n^{2}-13n+36 =0\\\\\\Teorema \ Vieta:\\\\n_{1} +n_{2} =13\\\\n_{1}\cdot n_{2} =36\\\\n_{1} =4\\\\n_{2} =9\\\\\\Otvet: \ 2 \ ; \ 4 \ ; \ 9$