In my book
for this question:
How many ways are there to split a dozen people into 3 teams, where each team has 4 people?
The answer is:
There are $\frac{12!}{4!4!4!}$ ways to divide the people into a Team A, a Team B, and a Team C, if we care about which team is which (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 $\frac{12!}{4!4!4!3!} = 5775$
But then this question:
A group of 30 dice are thrown. What is the probability that 5 of each of the values 1, 2, 3, 4, 5, 6 appear?
Gives the number of favorable events as $\frac{30!}{(5!)^6}$ because it is the number of permutations of the sequence 1111122222...66666.
I know that the numbers can be rearranged in 30! possible ways and then we adjust for overcounting the number of ways that each group of 5 can be arranged within that group (the 1s can come in any order, so divide by 5! and the same for the other numbers), giving the $(5!)^6$.
However, I would think that (as in the first problem) we also don't care if the first sequence of 5 numbers is the 2's group or the 3's group etc. so shouldn't we divide by 6! as well (the number of ways in which the groups of 5 can be rearranged)? (for a final result of $\frac{30!}{(5!)^66!}$)