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?

  • 4
    $\begingroup$ 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

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.


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:


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).


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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