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Combination and Permutation questions

  1. Dec 15, 2004 #1
    Hi,
    Can anyone help me with these permutation/combination questions?

    Solve the equation for n:
    1. nC4 = 35
    2. nC4 = 70

    It would be really good if I got the answers with full explanations, a.s.a.p. Thanks.
     
    Last edited: Dec 15, 2004
  2. jcsd
  3. Dec 16, 2004 #2

    matt grime

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    THere is nothing wrong with trial and error some times.

    nCr is increasing (for a fixed r)

    Alternatively write nC4=35 out in factorials and get a 4th order polynomial to solve.

    You can improve this since it is the same as n(n-1)(n-2)(n-3)=35*24

    so n is very close to the 4th root of 35*24
     
  4. Jan 7, 2005 #3
    solve(n*(n-1)*(n-2)*(n-3)=35*24,n); -4, 7, 3/2 + 1/2 I sqrt(111), cc.

    solve(n*(n-1)*(n-2)*(n-3)=70*24,n); -5, 8, 3/2 + 1/2 I sqrt(159), cc

    I cannot remember the formula for quartic polynomials. Just look up in Abramowich or use Maple.
     
    Last edited: Jan 7, 2005
  5. Jan 7, 2005 #4

    HallsofIvy

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    Since n must be an integer, it's easier to use "trial and error". In particular 6C4= [itex]\frac{(6)(5)}{(2)}[/itex]= 15 which is too small while 7C4= [itex]\frac{(7)(6)(5)}{(3)(2)}[/itex]= 35. Aha!

    8C4= [itex]\frac{(8)(7)(6)(5)}{(4)(3)(2)}[/itex]= 70.
     
  6. Jun 29, 2008 #5
    yes
    nC4=n!/4!.(n-4)!=n(n-1)(n-2)(n-3)[(n-4)!]/(n-4)!.24=35
    now n(n-1)(n-2)(n-3)=35*24
    n(n-1)(n-2)(n-3)=7*6*5*4
    compairing both sides we gwt
    n=7
    n-1=6 so n=7
    ans n=7
     
  7. Jun 29, 2008 #6

    matt grime

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    This thread is 3.5 years old, doubt that it's going to be of interest to the OP now.
     
  8. Jun 29, 2008 #7

    HallsofIvy

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    I wondered why I didn't recognize my own response!
     
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