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Thread: Combinations average

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    Combinations average




    I have 6 averages, 365, 406, 420, 416, 434, 435. How many ways can you split the group to have two groups of 3 with similar averages? The closer the average, the better. Each average can only be in one group.

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    Re: Combinations average


    This will depend on how similar should similar groups be to be considered similar ? You have C(6,2) =6!/4!*2!=15 different groupings, the similarity in each should be easy to calculate.

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