In how many ways can four prizes be distributed to three persons if every person receives at least one prize?

combinatorics

A possibly easier solution: Temporarily imagine a fourth person, and give everybody one prize. This can be done in $4!=24$ ways. Now disqualify the fourth person and award his prize to one of the others. This can be done in $3$ ways, as you have $3$ others to give the prize to. You now have the situation in the question, but each distribution has been counted twice, because the person with two prizes could have had either of those prizes initially and gotten the other one from the disqualified person. So we divide by $2$ to correct the overcount. We get $24\times3/2=36$.

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