Question

Sendo E = A5,0 + A6, 2 F = $\frac{A_{5,3}+A_{4,4} }{3}$, o valor de E + F é

a) 7

b) 31

c) 24

d) 59

e) 217

Gabarito Plurall: d)

Answer 1

Arranjo:

$A_{n,p} = \dfrac{n!}{(n-p)!}$

$A_{5,0} = \dfrac{5!}{(5-0)!} = \dfrac{5!}{5!} = 1$

$A_{6,2} = \dfrac{6!}{(6-2)!} = \dfrac{6!}{4!} = 6 \cdot 5 = 30$

Assim, podemos calcular E:

$E = A_{5,0} + A_{6,2} = 1 + 30 = 31$

$A_{5,3} = \dfrac{5!}{(5-3)!} = \dfrac{5!}{2!} = 5 \cdot 4 \cdot 3 = 60$

$A_{4,4} = \dfrac{4!}{(4-4)!} = \dfrac{4!}{0!} = 4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$

Assim, podemos calcular F:

$F = \dfrac{60 + 24}{3} = \dfrac{84}{3} = 28$

Logo:

$E + F = 31 + 28 = 59$

Alternativa D