Question

alguém me ajuda a resolver essas questões de combinaçoes simples???

Answer 1

Resposta:

A) 36

B) 10

C) 12

D) 4

Explicação passo-a-passo:

$C_{n,p} = \frac{n!}{p!(n-p!)}$

A) $C_{9,2} = \frac{9!}{2!*7!} = \frac{9*8*7!}{2!*7!} = \frac{9*8}{2*1} = \frac{72}{2} = 36$

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B) $C_{5,3} = \frac{5!}{3!*2!} = \frac{5*4*3!}{3!*2!} = \frac{5*4}{2*1} = \frac{20}{2} = 10$

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C) $C_{10,3} = \frac{10!}{3!*7!} = \frac{10*9*8*7!}{3!*7!} = \frac{10*9*8}{3*2*1} = \frac{720}{6} = 120\\C_{5,3} = \frac{5!}{3!*2!} = \frac{5*4*3!}{3!*2!} = \frac{5*4}{2*1} = \frac{20}{2} = 10$

$\frac{C_{10,3}}{C_{5,3}} = \frac{120}{10} = 12$

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D) $C_{n,2} = 6\\\frac{n!}{2!*(n-2)!} = 6\\\frac{n(n-1)(n-2)!}{2!(n-2)!} = 6\\\frac{n(n-1)}{2!} = 6\\n^2-n=6*2\\n^2-n-12=0\\n = \frac{-b +- \sqrt{b^2-4ac} }{2a}\\n = \frac{1 +- \sqrt{1+48}}{2}\\n = \frac{1 +- 7}{2} = \frac{1+7}{2}, \frac{1-7}{2}$

Desconsidera o: $\frac{1-7}{2}$

$n = \frac{1+7}{2} = \frac{8}{2} = 4$