In how many ways can nine students be divided into three labeled teams (A, B, C) of three people

Question

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.

Answer 1

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.