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: Permutations Problem

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve for n.

    [tex](14)_{n}P_{3}=_{n+2}P_{4}[/tex]

    2. Relevant equations
    [tex]_{n}P_{r}=\frac{n!}{(n-r)!}[/tex]

    3. The attempt at a solution

    First I write the problem with the equations written out

    [tex]\frac{14n!}{(n-3)!}=\frac{(n+2)!}{(n-2)!}[/tex]

    I'm not quite sure how to isolate the n with all of the factorials going on.
    If I subtract the RHS over to the left I get:

    [tex]\frac{(n-2)!14n!-(n+2)!(n-3)!}{(n-2)!(n-3)!}[/tex]

    Not sure if this is helpful or not.
     
    Last edited: Jul 11, 2010
  2. jcsd
  3. Jul 11, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Expand the factorials out a bit and cancel terms. For example,

    [tex]\frac{(n+2)!}{(n-1)!} = \frac{(n+2)(n+1)n(n-1)!}{(n-1)!} = (n+2)(n+1)n[/tex]
     
  4. Jul 11, 2010 #3
    Do you have (n-1)! in the denominator by accident? because it really comes out to be (n-2)!.

    Ok, I see your method,

    [tex]\frac{14(n-3)!(n-2)(n-1)(n)}{(n-3)!}=\frac{(n-2)!(n-1)(n)(n+1)(n+2)}{(n-2)!}[/tex]

    Some canceling and simplifying and you'll get

    [tex]0=\frac{(n+1)(n+2)}{14(n-1)}[/tex]

    and that dosen't divide nicely at all. Am I doing something wrong?
     
  5. Jul 11, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    It was just an example; it had nothing to do with the problem.
    You don't need to divide. If the fraction is 0, that means the numerator is 0. What values of n make the top equal to 0?

    Actually, the LHS shouldn't be 0. Check your work. I think your denominator is off too.
     
  6. Jul 11, 2010 #5
    I originally had

    [tex]14(n-2)(n-1)(n)=(n-1)(n)(n+1)(n+2)[/tex]

    then I divided both sides by (n-2)(n-2)(n) to cancel terms and set one side equal to zero so I can solve for n.

    Your right about the denomonator though it should be (n-2)14
     
  7. Jul 11, 2010 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You can't just arbitrarily set one side equal to 0. If you just cancel the factors common to both sides, you get

    [tex]14(n-2) = (n+1)(n+2)[/tex]

    How would you solve that?
     
  8. Jul 11, 2010 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes, as long as you're not dividing by 0. I added to my previous post, so you may want to look at it again.
     
  9. Jul 11, 2010 #8
    Ahh Foil out and move to one side.

    so you get

    [tex]0=n^2-11n+30[/tex]
    [tex]0=(n-5)(n-6)[/tex]

    therefore n=5,6 and that's what the answer is supposed to be.

    Thanks!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook