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In how many ways can 8 students be assigned to 2 groups if each group must have at least 3 students?

  • Combinatorics -

    so the groups could be 3,5 ; 4,4 ; or 5,3

    for a total of C(8,3)xC(5,5) + C(8,4)xC(4,4) + C(8,5)xC(3,3)
    = 56 + 70 + 56 = 182

    I am assuming that the two groups can be distinguished.
    If not, then it would simply be 56+70

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