Question

A6,3 me ajudam é urgente

Answer 1

$A_{n,p}=\frac{n!}{(n-p)!}\\ \\A_{n, 0}=\frac{n!}{(n-0)!}=\frac{n!}{n!}=1\\\\A_{1, 1}=\frac{1!}{(1-1)!}=\frac{1!}{0!}=\frac{1}{1}=1\\\\A_{n, 1}=\frac{n!}{(n-1)!}=\frac{n\cdot(n-1)!}{(n-1)!}=n\\\\A_{6, 3}=\frac{6!}{(6-3)!}=\frac{6!}{3!}=\frac{6\cdot5\cdot4\cdot3!}{3!}=6\cdot5\cdot4=120$