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Thread: Number of elements in a summation set?

  1. #1

    Thread Starter
    pathfinder NotLKH's Avatar
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    Number of elements in a summation set?

    I was wondering if the attachment illustrating the development of the formula:

    [S!/{(S-Z1-Z2-Z3...!)*(Z1!)*(Z2!)*(Z3!)...}]

    is correct, or did I make a mistake somewhere?

    -Lou
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  2. #2
    Fanatic Member sql_lall's Avatar
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    Talking looks like it

    first off, the equation itself is right
    (is essentially the same as picking, from the S objects, separate sets of Z1, Z2, Z3, ....S-Z1-Z2-Z3-....) - note that the last set is all elements which will be raised to the power of zero


    And splitting it up like that looks perfectly fine...It's just like saying 'Pick Z1 objects from S, then Z2 from (S-Z1), then...'

    So yeah, looks all fine to me
    Certainly makes sense, if that's what ur asking.

    Then only thing you might have to say is that none of the Ys can be zero...cos then it may as well not be there
    sql_lall

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