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