Question

Simplifique o arranjo:
$\sf\dfrac{A_{8,2}+A_{4,3}-A_{5,3}}{A_{10,3}-A_{9,1}}$​

Answer 1

$A_{8,2} = \cfrac{8!}{6!} = 8 \cdot 7 = 56$

$A_{4,3} = \cfrac{4!}{1!} = 4! = 24$

$A_{5,3} = \cfrac{5!}{2!} = 5 \cdot 4 \cdot 3 = 60$

$A_{10,3} = \cfrac{10!}{7!} = 10 \cdot 9 \cdot 8 = 720$

$A_{9,1} = \cfrac{9!}{8!} = 9$

$\cfrac{A_{8,2} + A_{4,3} - A_{5,3}}{A_{10,3} - A_{9,1}} \\\\\\= \cfrac{56 + 24 - 60}{720 - 9}\\\\\\= \cfrac{20}{711}$