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    The digits 12345 can be rearranged in 5! = 120 different ways.

    But because 99984 contains three copies of 9, the 120 possibilities lead to multiple counting.

    Since the three 9's can themselves be arranged in 3! = 6 ways, you need to divide the 120 by 6 to give 20.
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    You'd start off with the same 120 ways.

    Then you'd divide by 2 = 2! for the double counting of ways of arranging the duplicate 9's
    and then divide by another 2 = 2! for double counting of 8's

    so 30
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    (Original post by Farhan.Hanif93)
    Does this process have a name?
    Not that I know of.
 
 
 
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