Probability in a Restaurant In a revolving restaurant, there are four round tables each with three seats. How many different ways can $12$ people sit in this restaurant?
This is what I think the answer is:
$$\binom{12}{4} \cdot 3 \cdot 2 \cdot 1 \cdot \frac{1}{3} \cdot \binom43$$
Is this right?

I assume that $(abc)(def)(ghi)(jkl)$ is considered the same result as $(def)(ghi)(jkl)(abc)$ (i.e. rotations of tables is irrelevant) and $(bca)(def)(igh)(klj)$ (i.e. rotations within tables is irrelevant), but is considered different from $(bac)(def)(ghi)(jkl)$ (mirrors of tables considered different) and is considered different from $(def)(abc)(ghi)(jkl)$ (tables must remain in same order).
First, with the tables stopped still, assign each person to a table.  There are $\frac{12!}{3!3!3!3!}$ ways of doing that.
For each table, arbitrarily select one person to be special (by height or whiteness of teeth or name, it doesn't matter) and choose which of the two other people at the table sits on his/her left.
Finally, start the tables rotating and "forget" which table was situated in the north at the beginning (divide by 4)
The total number of ways then is $\frac{12!}{3!3!3!3!}\cdot 2^4\cdot \frac{1}{4}$

An equivalent solution relying more heavily on "dividing by symmetry"
Arrange everyone in a line: $12!$ number of ways to do this
The first three people in the line send to the first table and have them seat in the chairs in clockwise order as they had appeared in line starting from the north.  Do so similarly for the next group of three and the next group of three, etc...
"Forget" who was sat at the north on each table (divide by $3$ once for each table)
"Forget" which table was at the north to begin with once the restaurant starts spinning (divide by $4$)
For a total of $12!\cdot\frac{1}{3^4}\cdot \frac{1}{4}$ for the same answer as before.
