Question

arranjos simples com fatoriais:A6,3+A5,3-A3,2

Answer 1

Vamos usar essa formula aqui: $A_{n,p} = \dfrac{n!}{(n - p)!}$

$A_{6,3} = \dfrac{6!}{(6 - 3)!} = A_{6,3} = \dfrac{6 \cdot 5 \cdot 4 \cdot 3!}{3!}$

$A_{6,3} = 6 \cdot 5 \cdot 4 = 120$

$A_{5,3} = \dfrac{5!}{(5 - 3)!}$

$A_{5,3} = \dfrac{5!}{2!} = A_{5,3} = \dfrac{5 \cdot 4 \cdot 3 \cdot 2!}{2!}$

$A_{5,3} = 5 \cdot 4 \cdot 3 = 60$


$A_{3,2} = \dfrac{3!}{(3 - 2)!}$


$A_{3,2} = \dfrac{3!}{1!} = A_{3,2} = \dfrac{3 \cdot 2 \cdot 1!}{1!}$

$A_{5,3} = 3 \cdot 2 = 6$

$A_{6,3} + A_{5,3} - A_{3,2} = 120 + 60 -6$

$\boxed{A_{6,3} + A_{5,3} - A_{3,2} = 174}$

Espero ter te ajudado!