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anonymous

  • one year ago

solve for n P(n,3) = 60

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  1. Luigi0210
    • one year ago
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    I'm assuming the P means Permutations.. do you know the formula?

  2. anonymous
    • one year ago
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    yes

  3. anonymous
    • one year ago
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    P(n,r) = n!/(n-r)!

  4. Luigi0210
    • one year ago
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    So I'm assuming they want us to work backwards.. \(\large P(n, r ) = 60 = \frac{n!}{(n-3)!}\) And we know that that \(n> 3 \)

  5. anonymous
    • one year ago
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    yeah, for sure :)

  6. Luigi0210
    • one year ago
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    I don't think they'd make the problem too difficult.. so I'd suggest just plugging in values for n that are greater than 3. So if we try 4; 4! would equal 24, which is too small and not equal to 60. So 4 won't work. Now trying 5: \(\large P(5, 3) = \frac{5!}{(5-3)!} = \frac{120}{2!}\) And what's 120/2?

  7. anonymous
    • one year ago
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    how to solve this then? i'm required to show a solution

  8. anonymous
    • one year ago
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    \[0=n ^{3}-3n ^{2}+2n-60\]

  9. anonymous
    • one year ago
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    i have went all until this equation, and i'm stuck

  10. anonymous
    • one year ago
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    but 5's gonna work for that equation for sure

  11. Luigi0210
    • one year ago
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    I'm not sure if they want you to go into it that deep, I think just plugging in values would suffice

  12. anonymous
    • one year ago
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    yeah they do! and it's HORRIBLE

  13. anonymous
    • one year ago
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    but thanks anyway :)

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