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Permutations of n taken r at a time

  1. Jan 21, 2004 #1
    How can I solve these two problems?

    P(n,3)=210 and P(5,r)=20

    For the first one I got up to n(n-1)(n-2)=210 but I dont know how to solve a cubic equation...And the second one I have no clue. I'd appreciate some help, thanks
     
  2. jcsd
  3. Jan 21, 2004 #2

    Hurkyl

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    Trial and error; there are very few possibilities. :smile:
     
  4. Jan 21, 2004 #3
    Wow, its that simple? Is there any way to calculate an answer for a cubic equation?
     
  5. Jan 21, 2004 #4

    selfAdjoint

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    There's a general formula for the cubic equation, and also for the quadric equation. On the contrary Abel showed there is no general formula to solve the quintic equation (or higher) and Galois showed why, inventing group theory in the process.
     
  6. Jan 21, 2004 #5
    I see, thanks
     
  7. Jan 22, 2004 #6

    HallsofIvy

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    You can "speed up" the trial-and-error process a little by noting that, since n, n-1, and n-2 are about the same, the problem is "close to" n3= 210. 53= 125 and 63= 216 so start trying with n-1= 6.
    As for P(5,r)= 20, you know that r must be less than 5 so: P(5,0)= 1 (Of course, didn't really need to try that), P(5,1)= 5 (again "of course"), P(5,2)= 20!!!!!
     
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