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

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.