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Homework Help: Standard form permutation

  1. Feb 18, 2010 #1
    1. The problem statement, all variables and given/known data
    the problem states solve for n.
    nP4 = 8(nP4)


    2. Relevant equations

    no relevant equat. i can think of?

    3. The attempt at a solution
    my attempt at this was nP4 = 8(nP3)
    idk what i tried to do, but i tried to get it in standard form i guess: n!/ (n-4) = 8(n!)/(n-3)
    it doesnt seem right so im quite stuck. help please?
     
  2. jcsd
  3. Feb 18, 2010 #2

    Mentallic

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    Re: permutations

    No, you are on the right track. However you forgot the extra factorial signs:

    [tex]\frac{n!}{(n-4)!}=\frac{8n!}{(n-3)!}[/tex]

    Now solve for n by using what you know about factorials to simplify the equation.
     
  4. Feb 18, 2010 #3
    Re: permutations

    yeah, this is the stage im talking about, exactly what am i suppose to do to simplify.
    im thinking to cross mult. b/c its a proportion and turn it into a quad form? the problem is, the factorials; how can I simplify with all of them? and there are no factorials for N! (that I can solve out) :confused:
     
  5. Feb 18, 2010 #4

    Mentallic

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    Re: permutations

    Use the fact that [itex]n!=n(n-1)!=n(n-1)(n-2)![/itex] etc. and solve the equation like you would any other equation (dividing by n! on both sides would be a good start).
     
    Last edited: Feb 18, 2010
  6. Feb 18, 2010 #5
    Re: permutations

    hmm okay i understand.
    i tried that and this was my process/answer:
    i cross mult - n!(n-3)!=8n!(n-4)
    i then divided (as you suggested) n!(n-3)!/8n! = 8n!(n-4)!/8n!
    i then tried to isolate the N by adding the 4 onto the other side giving me (n+1)!/8 = n!

    I tried solving like a normal equation by dividing but the N on both sides confuses me, if I do something with it; won't it cancel?
    what step did I get wrong?
     
  7. Feb 18, 2010 #6

    Mentallic

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    Re: permutations

    You're making things more complicated than they need to be. Let's start again.

    [tex]\frac{n!}{(n-4)!}=\frac{8n!}{(n-3)!}[/tex]

    now, using the rule I told you, we will apply it in a way to make both sides more equivalent in a sense.
    Notice that if [itex]n!=n(n-1)![/itex] then similarly, [itex](n-3)!=(n-3)(n-4)![/itex]

    So we have [tex]\frac{n!}{(n-4)!}=\frac{8n!}{(n-3)(n-4)!}[/tex]

    Now divide both sides by [tex]\frac{n!}{(n-4)!}[/tex] and you're cleared of all factorials.
     
  8. Feb 18, 2010 #7

    Mentallic

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    Re: permutations

    Oh and by the way, you went wrong on this line:

    From the line before: [tex]\frac{n!(n-3)!}{8n!} = \frac{8n!(n-4)!}{8n!}[/tex]

    Simplified: [tex]\frac{(n-3)!}{8}=(n-4)![/tex]

    And now using the rule (properly) you should have [tex]\frac{(n-3)(n-4)!}{8}=(n-4)![/tex]

    And now you can divide through by (n-4)!
     
  9. Feb 21, 2010 #8
    Re: permutations

    ooh okay thanks :)
    you helped me alot, there are some aspects i still dont understand however, but i'll ask my teacher. youre a big help.
     
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