Vtyjkyuk

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:
$$
12 choose 2 10 choose2 8choose26choose33choose3frac12!2!2!frac13!3!
$$

Although the correct solution should be :
$$
12 choose 2 10 choose2 8choose26choose33choose3frac12!frac13!
$$
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?

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
$$
fracbinom62 binom42 binom223!.
$$
Again, the $3!$ is coming from the number of groups, not their size.

The number of ways of chosing r objects from a collection of n, $^nC_r$, is $fracn!r!(1-r)!$
There are $^12C_6$ 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 $$frac^12C_6times2times^6C_3times^6C_2times^4C_23 = frac2times12!times6!times6!times4!3times6!times6!times3!times2!times2!=frac12!times4!3times2!=frac4times12!5$$

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:

$$binom63binom33.$$

For any given outcome, say $A,B,CD,E,F$, from this formula, all other permutations (in this case only $D,E,FA,B,C$) 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 $binom64binom22$, and $binom21binom21$ (which is the case of choosing an apple out of two fruits and an orange out of two fruits).