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Thread: Splitting sum of permutation

  1. #1
    Senior Member oldguynewstudent's Avatar
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    Splitting sum of permutation

    OK, would the following be correct?

    $\displaystyle \sum_{j=0}^{k+1}(k+1)_{j}$=

    $\displaystyle (k+1)[k! + \sum_{j=0}^{k}(k)_{j}]$

    where $\displaystyle (k)_j = _kP_j$
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  2. #2
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    Quote Originally Posted by oldguynewstudent View Post
    OK, would the following be correct?

    $\displaystyle \sum_{j=0}^{k+1}(k+1)_{j}$=

    $\displaystyle (k+1)[k! + \sum_{j=0}^{k}(k)_{j}]$

    where $\displaystyle (k)_j = _kP_j$
    Dear oldguynewstudent,

    When k=1,

    $\displaystyle \sum^{k+1}_{j=0}{(k+1)_{j}}=\sum^{k+1}_{j=0}{~^{k+ 1}P_{j}}=5$

    $\displaystyle (k+1)\left[k! + \sum_{j=0}^{k}(k)_{j}\right]=6$

    Therefore, $\displaystyle \sum^{k+1}_{j=0}{(k+1)_{j}}\neq(k+1)\left[k! + \sum_{j=0}^{k}(k)_{j}\right]$

    Hope this will help you.
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