Question

Calcule:
a) A7,3 - 2.C 6,2=

b) 6! - 8! - 9!/ SOBRE 6!

Answer 1

a)
$A_{7,3}-2 \cdot C_{6,2} \\\\ \frac{7!}{(7-3)!}-2 \cdot \frac{6!}{(6-2)! \cdot 2!} \\\\ \frac{7!}{4!}- \frac{2 \cdot 6!}{4! \cdot 2!} \\\\ \frac{7 \cdot 6 \cdot 5 \cdot \not 4!}{\not 4!}- \frac{\not 2 \cdot 6 \cdot 5 \cdot \not 4!}{\not 4! \cdot \not 2} \\\\ 7 \cdot 6 \cdot 5 \cdot 6 \cdot 5 = \boxed{\boxed{6.300}}$


b)
$\frac{6!-8!-9!}{6!} = \frac{6!-8 \cdot 7 \cdot 6!-9 \cdot 8 \cdot 7 \cdot 6!}{6!} = \frac{\not 6! \cdot (1-8 \cdot 7-9 \cdot 8 \cdot 7)}{\not 6!} = 1-56-504 = \boxed{\boxed{-559}}$