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mixtapevanity
04-13-2008, 03:53 PM
In how many ways can I divide 18 students into groups of 3?

Originally I thought of it as 18 choose 3, where order doesn't matter, so then we have 15 remaining. So we do 15 choose 3, divide it by 3! since order doesn't matter and continue until we are at 3 student left, which means only one choice. Then multiply those numbers together and I get something like 1.37225088 * 10^11.

Is this correct?

pka
04-13-2008, 05:07 PM
Provided that the groups are not identifiable, that is say study groups as opposed to teams with different names, then you are asked for unordered partitions.
\frac{18!}{[(3!)^6](6!)}.

galactus
04-14-2008, 06:01 PM
You must divide by 6! instead of 3!.

\frac{\Pi_{k=0}^{5}C(18-3k,3)}{6!}