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Homework Help: Factorial Notation

  1. Apr 2, 2006 #1
    Hey, I was wondering if someone could help me with a specific type of question that I can't seem to understand without an answer key. Anyway, it's rewriting expressions with factorial notation so that they no longer have factorial symbols.

    Example:

    Simplify without using the factorial symbol:
    (n-2)!(n+1)!/(n!)²

    The answer is: n+1/n(n-1)

    What I don't understand is how you come to that conclusion. Can someone explain this to me?
     
  2. jcsd
  3. Apr 2, 2006 #2
    well, the key to solution is to notice that n!=(n-2)!*(n-1)*n and (n+1)!=n!*(n+1). also (n!)^2=n!*n!.
     
  4. Apr 2, 2006 #3

    dav2008

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    What is n! ?

    It's n*(n-1)*(n-2)*(n-3)*...*1

    What is (n-2)! and (n+1)! ?

    If you write all of those out, you'll notice that certain terms cancel.

    Edit: Or I guess a more direct approach would be to write n! in terms of (n-2)! like mantito has done.
     
  5. Apr 2, 2006 #4
    Yeah, I understand what you're saying, but I'm still stuck.

    I guess I'll show you my work.

    (n-2)!(n+1)!/(n!)²

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

    = (n+1)!/n!

    What I have trouble with is I guess why the answer is (n+1)/n(n-1). Can I not just cancel out the factorial symbol without having to multiply (n-1)?
     
  6. Apr 2, 2006 #5

    nrqed

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    I don`t understand your numerator..did you use (n-2)!=(n-2)(n-1) n! ??
    That is incorrect!

    You just have to write the (n+1)! as (n+1) n! and then write one of the n! of the denominator as n (n-1) (n-2)! and then all the factorials will cancel out leaving you with (n+1) / (n (n-1))

    Patrick
     
  7. Apr 2, 2006 #6
    Thanks a bunch, but one more question: Why was mine incorrect?
     
  8. Apr 2, 2006 #7
    Because n! = (n-2)! (n-1) n
     
  9. Apr 3, 2006 #8

    VietDao29

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    The numerator in your second step is wrong!
    If it reads:
    (n - 2)! n! (n + 1) / (n (n - 1) (n - 2)! n!), then it's correct.
    Note that:
    [tex](n - 2)! \neq (n - 2) (n - 1) n![/tex]
    The LHS can be expanded as:
    (n - 2) (n - 3) (n - 4) ... 2 . 1
    Whereas the RHS is:
    n (n - 1)2 (n - 2)2 (n - 3) ... 2 . 1
    And of course the LHS is not equal the RHS, right?
    Can you get this? :)
     
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