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Elementary school math problem: three children, four different flowers, one for each, how many distribution methods are there?
If everyone has only one flower, it is: 4*3*2=24.

If one person spends two flowers, that is, two of the four (***C-4-2, that is, six kinds) are given to one person (there are three candidates) and the remaining two are given to the other two people (* * * two kinds of distribution), then the result is 6*3*2=36. . . PS: I don't think so.

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