How many ways $12$ persons may be divided into three groups of $4$ persons each?
I think the answer should be $\frac{12!}{(4!)^3}$ but the suggested correct answer is $5775$, could anybody explain where I am going wrong?
[Math] How many ways $12$ persons may be divided into three groups of $4$ persons each
combinatorics
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Best Answer
We can also organize the count in a different way. First line up the people, say in alphabetical order, or in student number order, or by height.
The first person in the lineup chooses the $3$ people (from the remaining $11$) who will be on her team. Then the first person in the lineup who was not chosen chooses the $3$ people (from the remaining $7$) who will be on her team. The double-rejects make up the third team.
The first person to choose has $\binom{11}{3}$ choices. For every choice she makes, the second person to choose has $\binom{7}{3}$ choices, for a total of $$\binom{11}{3}\binom{7}{3}.$$
Remark: The lineup is a device to avoid multiple-counting the divisions into teams. The alternate (and structurally nicer) strategy is to do deliberate multiple counting, and take care of that at the end by a suitable division.