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

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.
