Question

Como chegar a esse resultado?
-Arranjo Simples

Answer 1

$\dfrac{A_{n-1,3}}{A_{n,3}} = \dfrac{3}{4} \\\\\\ \dfrac{\frac{(n-1)!}{(n-4)!}}{\frac{n!}{(n-3)!}} = \dfrac{3}{4} \\\\\\ \dfrac{(n-1)! \cdot (n-3)!}{n! \cdot (n-4)!} = \dfrac{3}{4} \\\\\\ \dfrac{(n-1)! \cdot (n-3) \cdot (n-4)!}{n \cdot (n-1)! \cdot (n-4)!} = \dfrac{3}{4} \\\\\\ \dfrac{(n-3)}{n} = \dfrac{3}{4}$

$3 \cdot n = 4 \cdot (n-3) \\\\ 3n = 4n-12 \\\\ \boxed{\boxed{n = 12}}$

Temos que a solução é n valendo 12.