Question

Preciso de ajuda com análise combinatória fatorial:

a) 8!5! / 7!6! =
b) 8!+5! / 5! =
c) (n+2) / n!+(n+1)! =
d) (n+1)!+(n+2)! / (n+2)!n(n+1)! =
e) (n+1)!+(n+2)! / n! =

Answer 1

a)
$\boxed{\frac{8!5!}{7!6!}=\frac{8.7!5!}{7!6.5!}=\frac{8}{6}=\frac{4}{3}}$
b)
$\boxed{\frac{8!+5!}{5!}=\frac{8!}{5!}+\frac{5!}{5!}=\frac{8.7.6.5!}{5!}+1=337}$
c)
$\frac{(n+2)!}{n!+(n+1)!}=\frac{(n+2)!}{n!+(n+1)n!}=\frac{(n+2)!}{n!(n+2)}=\frac{(n+2)(n+1)!}{n!(n+2)} \\ \\ \frac{(n+1)!}{n!}=\frac{(n+1)n!}{n!}=n+1$
d)
$\frac{(n+1)!+(n+2)!}{(n+2)!n(n+1)!}=\frac{(n+1)!}{(n+2)!n(n+1)!}+\frac{(n+2)!}{(n+2)!n(n+1)!}= \\ \\ \frac{1}{(n+2)!n}+\frac{1}{n(n+1)!}$
e)
$\frac{(n+1)!+(n+2)!}{n!}=\frac{(n+1)!}{n!}+\frac{(n+2)!}{n!}=\frac{(n+1)n!}{n!}+\frac{(n+2)(n+1)n!}{n!}= \\ \\ (n+1)+(n+2)(n+1)=(n+1)(n+3)$