I've seen questions where it will have you arrange $N$ people at $2$ tables with $N\over 2$ people sitting at them. The answer is usually
$$\binom{N} {\frac{N}{2}} \cdot \left(\frac{N}{2} – 1\right)! \cdot \left(\frac{N}{2} – 1\right)!$$
Because with two tables having half the total number of people at each, we know what people are at both tables just by knowing the people at $1$, then we just multiply by all possible arrangements at each.
In a situation where by knowing one tables people we still do not know the others ($3$ tables or more) do we have to multiply by another $\binom{N}{K}$?
For example with $15$ people at $3$ different tables each seating $5$ people, is the number of combinations of seating equal to:
$$\binom{15}{5} \binom{10}{5}(4!)^3$$
And could we take this answer and divide by 3! to eliminate all situations where all arrangements are the same but just at different tables?
Best Answer
I think your computation is correct.
Here is an alternate way to approach this problem that I think is a little simpler.
For the two-table case: If you number all the seats, then there are $N!$ ways to seat everyone. Then, we divide by $(N/2)^2$ to account for overcounting due to symmetry at each table (we do not consider rotations of an arrangement at a table to be different). Note that this is the same as your computation.
$$\binom{N}{N/2}(N/2-1)! (N/2-1)! = \frac{N!}{(N/2)! (N/2)!}(N/2-1)! (N/2-1)! = \frac{N!}{(N/2)^2}$$
Note that I assume that the two tables are distinguishable; if they are not, just divide everything by $2$.
For your three-table example, we can do the same thing: there are $15!$ ways to seat everyone if we label all the chairs, and then we need to divide by $5^3$ to account for the symmetry at each table. This gives the same number as your computation.
$$\binom{15}{5} \binom{10}{5} (4!)^3 = \frac{15!}{5! 10!} \frac{10!}{5!5!} (4!)^3 = \frac{15!}{(5!)^3}(4!)^3 = \frac{15!}{5^3}.$$
Again, this computation is if the tables are distinguishable; otherwise, divide everything by $3!$.