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.
Is this what you are after? I am not sure so I will leave out the explanation and references until I am.
[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25], [26, 27, 28, 29, 30]]
[[1, 6, 11, 16, 21], [2, 7, 13, 19, 30], [3, 8, 14, 24, 29], [4, 9, 20, 25, 28], [5, 15, 18, 23, 27], [10, 12, 17, 22, 26]]
[[1, 8, 13, 20, 22], [2, 6, 12, 23, 28], [3, 9, 11, 18, 30], [4, 15, 16, 24, 26], [5, 7, 17, 25, 29], [10, 14, 19, 21, 27]]
[[1, 7, 18, 24, 28], [2, 10, 11, 20, 29], [3, 6, 15, 19, 22], [4, 14, 17, 23, 30], [5, 9, 13, 21, 26], [8, 12, 16, 25, 27]]
[[1, 10, 15, 25, 30], [2, 9, 17, 24, 27], [3, 7, 12, 20, 21], [4, 6, 13, 18, 29], [5, 14, 16, 22, 28], [8, 11, 19, 23, 26]]
[[1, 9, 12, 19, 29], [5, 6, 20, 24, 30], [4, 7, 11, 22, 27], [8, 15, 17, 21, 28], [3, 10, 13, 16, 23], [2, 14, 18, 25, 26]]
Best Answer
Because the groups of $2$ and $1$ are distinct sizes, you don't have to consider reordering those groups. The denominator should be $2!=2$ because there are $2$ ways to order the two groups of three. The numerator is correct.