Since the question explicitly says that all seats should be taken, and since those are circular tables I also made the assumption that orientations clockwise and anticlockwise are different, thus $(n-1)!$.
I ended up with the following solution:
$1^\circ$ Table : $\frac{15!}{4!}$
$2^\circ$ Table : $\frac{11!}{4!}$
$3^\circ$ Table : $\frac{7!}{4!}$
$4^\circ$ Table : $3!$
Number of possible combinations $ = \frac{15!}{4!} + \frac{11!}{4!} + \frac{7!}{4!} + 3! $
Would this be the correct way of solving the problem?
Best Answer
First, each partition of the $16$ people into groups of $4$ represents a different seating. The number of such partitions is
$$P = \binom{16}{4} \times \binom{12}{4} \times \binom{8}{4} = \frac{(16!)}{(4!)^4}.$$
Note that the above computation accommodates that the $(4)$ tables are to be considered distinguishable from each other. If that were not the case, then you would have to add the factor of $\displaystyle \frac{1}{(4!)}$ to represent that each partitioning would otherwise be overcounted $(4!)$ times.
Then, for each such partition, since a specific table is round, you can arbitrarily assign one person at the head of the table, and then reason that the remaining $(3)$ people can be seated in $(3!)$ different ways around the table.
Therefore, the final computation is
$$P \times (3!)^4 = \frac{(16!)}{(4!)^4} \times (3!)^4 = \frac{(16!)}{(4)^4}.$$