Question

me ajudem em algumas dessas tarefas

Answer 1

$\\ \\ 1-a)= \frac{7!}{4!} = \frac{7.6.5.4!}{4!} =7.6.5 = 210 \\ \\ 1-b)= \frac{3!.7!}{4!.6!} =\frac{7!.3!}{6!.4!}= \frac{7.6!.3!}{6!.4.3!} = \frac{7}{4} =1,75 \\ \\ 2-a)= \frac{n!}{(n-1)!} =\frac{n!}{(n-1)n!} = \frac{1}{(n-1)} \\ \\ 2-b)= \frac{(n+2)!}{n!} =\frac{(n+2)(n+1)n!}{n!}=(n+2)(n+1)= n^{2}+3n+2 \\ \\ 3- \frac{(n+1)!}{n!}= 4 \\ \\ \frac{(n+1)n!}{n!}= 4 \\ \\ n+1= 4 \\ \\ n = 4 - 1 \\ \\ n = 3 \\ \\ 4-a)= \frac{n!}{(n-2)!} =30[tex]$ \frac{n!}{(n-2).(n-1)n!}= \frac{1}{}