In how many ways can nine students be divided into three labeled teams (A, B, C) of three people?
My answer, which I am not sure if correct, is $\frac{9!}{3!3!3!}$ ways.
I am not sure if there are equivalent set of groups just like what we need to consider for unlabeled groups case with the same given grouping condition.
Any comments and suggestions will be much appreciated. Thank you in advance.
On
There are $\dfrac{9!}{6!3!}$ ways to place $3$ students in the first group, and with the remaining $6$ students, there are $\dfrac{6!}{3!3!}$ ways to place $3$ students in the second group, and the remaining $3$ students will be placed in the third group. Therefore, there are in all $\dfrac{9!}{6!3!}\times\dfrac{6!}{3!3!}=\dfrac{9!}{3!3!3!}$ ways to put $9$ students into $3$ groups of $3$.
That looks correct.
First choose team A: ${9 \choose 3}$ ways.
Then choose team B from who's left: ${6 \choose 3}$ ways.
Then team C is chosen already.
So,
$$\frac{9!}{6!3!}\frac{6!}{3!3!} = \frac{9!}{3!3!3!}$$ ways, just like you found.
If the groups are unlabeled, then we've counted $3!$ times too many arrangements (because we can switch the team labels). So we divide by another $3!$ for the unlabeled case.