Question

Aranjamente de n elemente luate câte 7 - aranjamente de n elemente luate câte 5 supra aranjamente de n elemente luate câte 5 = 89

Answer 1

 

$\displaystyle\bf\\ \text{Folosim formula: }~~A_n^k=\frac{n!}{(n-k)!}\\\\ \text{Rezolvare: }\\\\\frac{A_n^7-A_n^5}{A_n^5}=89\\\\\\\frac{A_n^7}{A_n^5}-\frac{A_n^5}{A_n^5}=89\\\\\\\frac{A_n^7}{A_n^5}-1=89\\\\\\\frac{A_n^7}{A_n^5}=89+1\\\\\\\frac{A_n^7}{A_n^5}=90\\\\\\\frac{~~\dfrac{n!}{(n-7)!}~~}{\dfrac{n!}{(n-5)!}}=90\\\\\\\\\frac{n!}{(n-7)!}\times\frac{(n-5)!}{n!}=90\\\\\\\frac{(n-5)!}{(n-7)!}=90\\\\\\\frac{(n-7)!\times(n-6)(n-5)}{(n-7)!}=90\\\\\\(n-6)(n-5)=90\\\\n^2-11n+30=90\\\\n^2-11n-60=0$

$\displaystyle\bf\\n^2-11n-60=0\\\\n_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{11\pm\sqrt{121+240}}{2}=\frac{11\pm\sqrt{361}}{2}=\frac{11\pm19}{2}\\\\\text{Alegem doar solutia pozitiva.}\\\\n=\frac{11+19}{2}=\frac{30}{2}\\\\\boxed{\bf~n=15}$