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
Call one 22-size group A and the other B. Since, it matters which is which, then the number of ways to make the three groups are $$\binom{60}{16}\binom{44}{22}\binom{22}{22} = \binom{60}{16,22,22} = 3.147908\times 10^{26},$$ which is the same as the original answer. In other words, I choose 16 to be in the small group. Then I am free to choose 22 from the remaining to be in A, and the I am left with group B.