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How many ways to divide group of 12 people into 2 groups of 3 people and 3 groups of 2 people?

my answer to this question is: $$ {12 \choose 2}{ 10 \choose2 }{8\choose2}{6\choose3}{3\choose3}\frac{1}{2!2!2!}\frac{1}{3!3!} $$

Although the correct solution should be : $$ {12 \choose 2}{ 10 \choose2 }{8\choose2}{6\choose3}{3\choose3}\frac{1}{2!}\frac{1}{3!} $$ What am I missing here? If I have 2 groups of 3 , and 3 groups of 2, shouldn't I divide each group by its factorial in order to cancel the inner ordering of the group?

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    $\begingroup$ I think the question is designed to confuse students by using the numbers 2 and 3 in two different ways. If you find it confusing, think about this related problem: How many ways to divide a group of 58 people into 4 groups of 7 people and 6 groups of 5 people? $\endgroup$ – Srivatsan Sep 17 '11 at 18:00
  • $\begingroup$ Thanks guys, i can see my mistake clearly now! @Srivatsan, Austin Mohr $\endgroup$ – MichaelS Sep 17 '11 at 18:05
  • $\begingroup$ Zero and zero, respectively. $\endgroup$ – Alexander Gruber Sep 10 '13 at 18:13
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The fact that ordering does not matter within a group is already taken care of by the binomial coefficients. The additional $2!$ and $3!$ you see in the answer are taking care of the fact that the order in which the groups themselves were chosen also does not matter.

For example, if your two-person groups are $\{A, B\}$, $\{C, D\}$, and $\{E, F\}$, then the following arrangements are all the same:

$\{A, B\}$, $\{C, D\}$, $\{E, F\}$

$\{A, B\}$, $\{E, F\}$, $\{C, D\}$

$\{C, D\}$, $\{A, B\}$, $\{E, F\}$

$\{C, D\}$, $\{E, F\}$, $\{A, B\}$

$\{E, F\}$, $\{A, B\}$, $\{C, D\}$

$\{E, F\}$, $\{C, D\}$, $\{A, B\}$

Notice there are $3!$ such arrangements. When you just multiply your binomial coefficients together, however, these all get counted as distinct. Dividing by $3!$ collapses these all into a single arrangement.

To give another example with a better selection of numbers, suppose you want to arrange 6 people into three groups of two each. This would be given by $$ \frac{\binom{6}{2} \binom{4}{2} \binom{2}{2}}{3!}. $$ Again, the $3!$ is coming from the number of groups, not their size.

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When you are continuously choosing the objects from the same group, you may probably be permuting them. For example, consider the formula in your question:

$$\binom{6}{3}\binom{3}{3}.$$

For any given outcome, say $\{A,B,C\}\{D,E,F\}$, from this formula, all other permutations (in this case only $\{D,E,F\}\{A,B,C\}$) exists. So you are actually permuting them. Since they mean the same in your question, you have to divide it by $2!$.

i.e. both are of same size and are chosen from the same group, so we actually permuted them. Therefore we don't have to consider repetition for $\binom{6}{4}\binom{2}{2}$, and $\binom{2}{1}\binom{2}{1}$ (which is the case of choosing an apple out of two fruits and an orange out of two fruits).

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The number of ways of chosing r objects from a collection of n, $^nC_r$, is $\frac{n!}{r!(1-r)!}$ There are $^{12}C_6$ ways to divide into 2 groups of 6. Then $^6C_3$ ways to divide a group of 6 into 2 groups of 3, $^6C_2$ ways to split into a 2 and a 4 and $^4C_2$ ways to split each 4 into 2s but then 15 of the possibilities would be identical. So it is $$\frac{^{12}C_6\times2\times^6C_3\times^6C_2\times^4C_2}{3} = \frac{2\times12!\times6!\times6!\times4!}{3\times6!\times6!\times3!\times2!\times2!}=\frac{12!\times4!}{3\times2!}=\frac{4\times12!}{5}$$

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