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Probability solve the expression P(2n+4,3) = 2/3P(n+4,4) for n

  1. Nov 5, 2005 #1
    i just want to confirm if my anser is right....
    this is the problem:

    solve the expression for n Є N

    P(2n + 4, 3) = 2/3P(n+4, 4)

    this is my work:

    (2n +4)!/(n-3)! = (2/3(n+4)!)/(n-4)!
    -2(n-2)!/(n-3)! = -2/3(n-4)!/(n-4)!
    -2/(n-3)! = -2/3
    -2 = -2/3(n-3)
    -2 = -2n/3 +6/3
    -2 = -2n/3 +2
    -6 = -2n +2
    -8 = -2n
  2. jcsd
  3. Nov 5, 2005 #2
    if by P(x,y) you mean # of permutations of x objects taken y at a time,
    so the formula for P(x,y) = x!/(x-y)!
    then no, your solution doesn't work.
    P(2n+4,3) = (2n+4)!/((2n+4) - 3)! = (2n+4)!/(2n+1)!
    same mistake on the other side of the equation.
    plus, you should at least say what P(x,y) is, otherwise
    no one knows what you're asking about.
  4. Nov 5, 2005 #3

    ok, i get what you mean, but how can i manipulate this, is seems i cannot cancel it... help again
    Last edited: Nov 5, 2005
  5. Nov 5, 2005 #4
    re: permutations

    this wat i did...

    (2n+4)!/((2n+4)-3)! = (2/3(n+4)!)/((n+4)-4)!
    (2n+4)!/(2n+1)! = (2/3(n+4)!)/(n)!

    i dont know wat do do next, im soo stuck, since you cannot cancel any of them, so should i cross multiply it?
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