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Math Help - Factorials

  1. #1
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    Factorials

    I understand how to find them when numbers are involved. However, my problem is I can not figure out how to do it with n terms.

    Example:

    \frac{(2n)!}{(2n-2)!}

    Any help will be greatly appreciated.
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by mathgeek777 View Post
    I understand how to find them when numbers are involved. However, my problem is I can not figure out how to do it with n terms.

    Example:

    \frac{(2n)!}{(2n-2)!}

    Any help will be greatly appreciated.
    Hint: \frac{(2n)!}{(2n-2)!} =  \frac{(2n)\times(2n-1)\times(2n-2)!}{(2n-2)!}
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  3. #3
    Kai
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    Hmm, try using the fact that n!= n(n-1)!=n(n-1)(n-2)!
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  4. #4
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    Quote Originally Posted by mathgeek777 View Post
    I understand how to find them when numbers are involved. However, my problem is I can not figure out how to do it with n terms.

    Example:

    \frac{(2n)!}{(2n-2)!}

    Any help will be greatly appreciated.
    \frac{(2n)!}{(2n-2)!}= \frac{1*2*...*(n-2)*(n-1)*n*(n+1)*(n+2)*...*(2n-2)*(2n-1)*(2n)}{1*2*...*(n-2)*(n-1)*n*(n+1)*(n+2)*...*(2n-2)}

    Do you see where you can cancel and reduce the expression?
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  5. #5
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    i dont know how sound this logic is since i haven't done factorials in a long time but maybe it will steer you in some sort of direction


    (2n)!
    ------
    (2n-2)!

    gives:

    1x2x3...x(2n-3) x(2n-2) x (2n-1) x (2n)
    --------------------------------------
    1x2x3...x((2n-2)-1) x (2n-2)

    this leaves (2n-1)(2n)
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  6. #6
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    So the answer is 2n(2n-1)?
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  7. #7
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    Quote Originally Posted by mathgeek777 View Post
    So the answer is 2n(2n-1)?
    Yes!
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  8. #8
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    try it out, test the values n = 3,4,5 first by inserting in the original expression and expanding and simplifying and then second by inserting the values n into 2n(2n-1) and you should get the same answer
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