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Math Help - Operations with factorials

  1. #1
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    Operations with factorials

    Hi, i'm having troubles understanding operations with faculty.

    If:

    (3k)! (k+1)! (2(k+1))!
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    k! (2k)! (3(k+1))!

    Then why is this equal to:

    (k+1)(2k+1)(2k+2)
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    (3k+1)(3k+2)(3k+3)

    Thanks for the help!
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  2. #2
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    Re: Operations with factorials

    Note that
    (3(k+1))!=(3k+3)!=(3k+3)(3k+2)(3k+1)(3k)!.
    And do it again with 2 instead of 3.
    Thanks from topsquark
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  3. #3
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    Re: Operations with factorials

    Hello, jones123!

    I must assume that you know how to "cancel" factorials.



    \text{Show that: }\: \frac{\big(3k\big)!\,\big(k+1\big)!\,\big(2[k+1]\big)!}{k!\,\big(2k\big)!\,\big(3[k+1]\big)!} \;=\;\frac{(k+1)(2k+1)(2k+2)}{(3k+1)(3k+2)(3k+3)}

    \text{We have: }\:\frac{(3k)!\,(k+1)!\,(2k+2)!}{k!\,(2k)!\,(3k+3)  !}

    . . . . . =\;\frac{(3k)!}{(3k+3)!}\cdot \frac{(k+1)!}{k!}\cdot\frac{(2k+2)!}{(2k)!}

    . . . . . =\;\frac{1}{(3k+3)(3k+2)(3k+1)}\cdot\frac{k+1}{1} \cdot\frac{(2k+2)(2k+1)}{1}

    . . . . . =\;\frac{(k+1)(2k+1)(2k+2)}{(3k+1)(3k+2)(3k+3)}


    But this reduces further . . .

    . . . . . =\;\frac{(k+1)\,(2k+1)\,2(k+1)}{(3k+1)\,(3k+2)\,3(  k+1)}

    . . . . . =\;\frac{2(k+1)(2k+1)}{3(3k+1)(3k+2)}
    Thanks from topsquark
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  4. #4
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    Re: Operations with factorials

    Thanks!
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