Question

Combinari de n luat cate 2 + aranjamente de (n-1) luat cate 2 <=32, aflati n

Answer 1

$C_{n}^2+A_{n-1}^2 \leq 32\\ \dfrac{n!}{(n-2)!\cdot 2!}+\dfrac{(n-1)!}{(n-3)!}\leq 32\\ \dfrac{(n-2)!\cdot (n-1)\cdot n}{(n-2)!\cdot 2}+\dfrac{(n-3)!\cdot (n-2)\cdot (n-1)}{(n-3)!}\leq 32\\ \dfrac{n(n-1)}{2}+(n-1)(n-2)\leq 32 |\cdot 2\\ n(n-1)+2(n-1)(n-2) \leq 64\\ n^2-n+2(n^2-n-2n+2)\leq 64\\ n^2-n+2(n^2-3n+2)\leq 64\\ n^2+2n^2-n-6n+4\leq 64\\ 3n^2-7n-60\leq 0\\ \Delta=49+4\cdot 3\cdot 60=769\\ n_1=\dfrac{7-\sqrt{769}}{6}\\ n_2=\dfrac{7+\sqrt{769}}{6}$

$\text{Tinand cont de faptul ca }n\in \mathbb{N},n\geq 3,\text{Obtinem:}\\ S:n\in \{3,4,5\}$