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Mathematics 20 Online
OpenStudy (anonymous):

in how many ways can 12 kids be placed on 3 distinct teams of 3,5, and 4 members?

OpenStudy (a_clan):

12C3 * 9C5 *4C4 Needs review

OpenStudy (zarkon):

one can also use the multinomial coefficient \[{12\choose3,5,4}=\frac{12!}{3!5!4!}\]

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