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Permutations Find the values ex. P(n,2)=90

  1. Sep 16, 2012 #1
    How can I solve this?

    Find the value of n.


    1. P(n,2)=90 ; 2. P(n,3)=3P(n,2)



    I tried like this but I am stuck

    1. n! / (n-2)! = 90 and then I don't know ...

    2. n!/(n-3!) = 3 n!/(n-2)! ⇔ n! (n-2)! / (n-3)! n! = 3 ⇔ (n-2)! / (n-3)! = 3 here the same


    As you can see I'm really stuck with this.
     
  2. jcsd
  3. Sep 16, 2012 #2

    HallsofIvy

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    You do understand that n!= n(n-1)(n-2)...(3)(2)(1), right?
    So (n-2)!= (n-2)...(3)(2)(1).
    [tex]\frac{n!}{(n-2)!}= \frac{n(n-1)(n-2)...(3)(2)(1)}{(n-2)...(3)(2)(1)}= 90[/tex]

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

    Cancel!
     
  4. Sep 17, 2012 #3
    Ok thanks, I think I understand what you mean,

    So 1) n(n-1)=90 then n=10
    2) n-2 = 3 then n=5

    but if I trie to solve this one it doesn't at up:

    3) 2P(n,2)+50=P(2n,2)

    2((n!/(n-2))!+50)=2n!/(2n-2)!

    Is this correct?
     
    Last edited by a moderator: Sep 17, 2012
  5. Sep 17, 2012 #4

    HallsofIvy

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    Not the way you have written it because you have the parentheses wrong. The last "!" should not be the whole "n!/(n-2)" and 2 does not multiply 50. On the right, it is clearer if you write (2n)! rather than 2n! which can be interpreted as 2(n!). What you should have is
    2(n!/(n-2)!+ 50= (2n!)/(2n-2)! or
    [tex]2\frac{n!}{(n-2)!}+ 50= \frac{(2n)!}{(2n-2)!}[/tex]

    Now write those factorials out in detail and cancel.
     
  6. Sep 17, 2012 #5
    I trie with Latex:

    [itex]2\frac{n!}{(n-2)! + 50 = \frac{2n!}{(2n-2)!}[itex\]
     
  7. Sep 17, 2012 #6
    ok, that didn't go the way I expected,hehe

    thank you HallsofIvy, now I have to learn how to use LaTeX..
     
  8. Sep 17, 2012 #7
    What do you mean by "and then cancel"???
     
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