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Math Help - Splitting sum of permutation

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

    OK, would the following be correct?

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

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

    where (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?

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

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

    where (k)_j = _kP_j
    Dear oldguynewstudent,

    When k=1,

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

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

    Therefore, \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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