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.
Best Answer
You thought right just a bit incomplete. The answer you got shows that that you divide seven students in
(7!)/(2!2!3!).
Suppose the 2 groups which have 2 people each is labeled as G1 and G2.
Now the labeling could have been G2 and G1 also.
There are 2! ways in which you can label the groups having 2 people each !
If there had been say 3 groups of a certain number of students where each group had 3 people. Then in the denominator you would have got
3!3!3!3! .
The extra 3! comes from the 6 possible arrangements of the 3 groups themselves.
Like say you have ABCDEF and you have to distribute this into 3 groups of 2 each.
One such group be
AB CD EF
Now this group can be also written as
CD EF AB , EF AB CD , CD AB EF , AB EF CD , EF CD AB. ( 3! =6 Ways )
Similarly for each possible combination of groups.
That's why you get an extra 2! ( or 3! ) term due to arrangement of the groups.