Question

Determine n, sabendo que A(n+1,4) = 20.C(n,2)

Answer 1

$A_{\,n+1\,,\,4}~=~20~.~C_{n,2}\\\\\\\frac{(n+1)!}{((n+1)-4)!}~=~20~.~\frac{n!}{2!.(n-2)!}\\\\\\\frac{(n+1)!}{(n-3)!}~=~20~.~\frac{n~.~(n-1)~.~(n-2)!}{2.(n-2)!}\\\\\\\frac{(n+1)~.~n~.~(n-1)~.~(n-2)~.~(n-3)!}{(n-3)!}~=~20~.~\frac{n~.~(n-1)~.~1}{2~.~1}\\\\\\\frac{(n+1)~.~n~.~(n-1)~.~(n-2)~.~1}{1}~=~10~.~n~.~(n-1)\\\\\\\frac{(n+1)~.~n~.~(n-1)~.~(n-2)~.~1}{n~.~(n-1)}~=~10\\\\\\(n+1)~.~(n-2)~=~10\\\\\\n^2-n-12~=~0\\\\Bhaskara\\\\\Delta~=~(-1)^2-4.1.(-12)~=~1+48~=~\boxed{49}$

$n'~=~\frac{1+\sqrt{49}}{2~.~1}~=~\frac{1+7}{2}~=~\boxed{4}\\\\n''~=~\frac{1-\sqrt{49}}{2~.~1}~=~\frac{1-7}{2}~=~\boxed{-3}$

Temos dois possíveis valores para n (4 e -3), no entanto precisamos lembrar que não existe fatorial de numero negativo, ou seja, n=-3 deve ser descartado.

Resposta: n = 4