1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook