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Math Help - Solving for permutation on one side of the equation and a combination on the other

  1. #1
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    Solving for permutation on one side of the equation and a combination on the other

    So far i've been asked to solve for n in either a Permutation or a Combination. I have a question that involves both. Any help with the following equation would be much appreciated (even a way to simplify it?).


    Here it is:

    Solve for n if P(n,3) = 2C(n,2), n belongs to N

    Thanks to anyone who replies.
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  2. #2
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    \frac{n!}{(n-3)!} = 2 \ \frac{n!}{2!(n-2)!}

     (n-2)! = (n-3)!

     (n-2)(n-3)! = (n-3)!

     n-2 = 1

     n=3
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  3. #3
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    I was just wondering what you divided by in the second line to come to (n-2)! = (n-3)!,

    I also have another question and you obviously know what you are doing:

    Solve for (n+2)!/n!=56

    You are awesome!
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  4. #4
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    Quote Originally Posted by jake.davis View Post
    I was just wondering what you divided by in the second line to come to (n-2)! = (n-3)!,

    Solve for (n+2)!/n!=56
    Note that the 2's divide off as well as the n!'s. Then the denominators must be equal.

    Now for the next problem note that: \frac{(n+2)!}{n!}=\frac{(n+2)(n+1)n!}{n!}
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  5. #5
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    Thanks Plato,
    For the second question the n!'s would cancel out leaving me with:

    (n+2)(n+1)=56
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