How many ways are there to split $12$ people into $4$ groups with $3$ people in each group? How many ways are there to split $12$ people into $4$ groups with $3$ people in each group?
I have tried to write all the possible ways to do it, and came up with $\frac{12!}{(3!)^3}\cdot4!$, however I am still wrong.
 A: 
I have tried to write all the possible ways to do it, and came up with $\frac{12!}{(3!)^3}\cdot4!$, however I am still wrong.

The answer you seek is $$\dfrac{12!}{(3!)^\mathbf 4\cdot 4!}$$
There are $12!$ ways to arrange 12 people in a line, and when split into 4 groups, there are $3!$ ways to arrange people in each group, and $4!$ ways to arrange the groups themselves. (Note: since the groups are all of the same size their place-holders are indistinguishable.)

For an easier example, there are $4!$ ways to arrange $\{a,b,c,d\}$ in a line, and when split into 2 groups, there are $2!$ ways to arrange letters in each group, and $2!$ ways to arrange the groups themselves. 
Since, for example, $\{\{a,b\},\{c,d\}\}$ is considered indistinguishable from $\{\{d,c\},\{b,a\}\}$, and other permutations within the groups and of the groups, we must count fewer (and therefore divide).
$$\dfrac{4!}{(2!)^2\cdot 2!}=3\\~\\ \{\{a,b\},\{c,d\}\}\\\{\{a,c\},\{b,d\}\}\\\{\{a,d\},\{b,c\}\}$$
A: One method may be choose 3 people from 12 then 3 from 9.....  $$\binom{12}3.\binom{9}3\binom{6}3.\binom{3}3.{1 \over 4!}$$
I didn't understood logic behind your attempt
A: The answer is 5775. There are (12! divided by (4!4!4!) ) ways to divide the people into a Team A, a Team B, and a Team C, if we care about which team is which (this is called a multinomial coefficient).
Since here it doesn’t matter which team is which, this overcounts by a factor of 3! , so the number of possibilities is 12! divided by $(4!4!4!3!) = 5775$
