How many ways $12$ persons may be divided into three groups of $4$ persons each?

We can also organize the count in a different way. First line up the people, say in alphabetical order, or in student number order, or by height.

The first person in the lineup chooses the $3$ people (from the remaining $11$) who will be on her team. Then the first person in the lineup who was not chosen chooses the $3$ people (from the remaining $7$) who will be on her team. The double-rejects make up the third team.

The first person to choose has $\binom{11}{3}$ choices. For every choice she makes, the second person to choose has $\binom{7}{3}$ choices, for a total of $$\binom{11}{3}\binom{7}{3}.$$

Remark: The lineup is a device to avoid multiple-counting the divisions into teams. The alternate (and structurally nicer) strategy is to do deliberate multiple counting, and take care of that at the end by a suitable division.

The answer is $\frac{12!}{(4!)^3\cdot3!}=5775$ because the $3!$ different orders of the three groups do not matter either, so your solution was almost correct.

From $12$ persons , $4$ persons can be chosen in $\binom{12}{4}$ ways. From rest $(12-4)=8 $ persons 4 persons can be chosen in $\binom{8}{4}$ ways.Remaining $4$ will form the third group.Thus $\binom{12}{4} \binom{8}{4}$ are the ways for form three groups of $4$ persons out of $12$ persons.But we have introduced order in group formation. Now three groups can be permuted $3!$ ways and they are all same.Hence correct number of ways $= \frac{\binom{12}{4} \binom{8}{4}}{3!}=5775 $