# Seating Multiple People at Multiple Tables

In how many ways can we seat 100 people around 20 different circular tables in such a way that there are five people per table?

Am I right in assuming that we're only considering unique situations?

For example, if we take 2 people at 2 tables with 1 per table. Then we'd have [A][B] and [B][A] which is actually only 1 arrangement. $\frac{C(2,1)}{2} = \frac{2}{2} = 1$

Or 4 people, 2 tables, 2 per {[AB] [CD]} {[AC] [BD]} {[AD] [BC]} which is three arrangements since we don't count [B,A], [C,A], [D,A], [D,C], [D,B] or [B,C] $\frac{C(4,2)}{2} = \frac{6}{2} = 3$

Therefore, the formula I got was $\frac{C(totalPeople,peoplePerTable)}{numberOfTables}$

For 100 people at 20 tables with 5 per table.

We have $\frac{C(100,5)}{20} = 3,764,376$ possibilities

## 2 Answers

There are $100!$ ways to line up the people. Take the first $5$ and seat them around the first table in order. Do the same for the second $5$ and the second table. Continue until you’ve seated all $100$. If the tables are individually identified, and each of the $20$ tables has a designated head seat, each of these $100!$ seatings counts as a different arrangement. If the tables are individually identified, but seatings that are equivalent under a rotation of the table are not considered distinct, then each seating at each table has been counted $5$ times, once for each possible location of a head seat, and there are therefore only $\dfrac{100!}{5^{20}}$ distinct arrangements.

You appear to be assuming, however, that the tables are not individually identified and that rotationally equivalent arrangements at a table are not distinguished. In that case each distinct arrangement has been counted $20!$ times in the figure $\dfrac{100!}{5^{20}}$, once for each of the $20!$ permutations of the tables, so there are only $\dfrac{100!}{5^{20}20!}$ distinct arrangements. This is a bit over $4\times10^{125}$.

This is actually an instance of a type of problem that can be solved by symbolic combinatorics for labelled counting. Let $\mathcal{Z}$ be the singleton class and $\mathcal{T}$ the set of all seating arrangements as required. This gives the combinatorial class specification $$\mathcal{T}= \mathfrak{P}_{20}(\mathfrak{C}_5(\mathcal{Z}))$$ where $\mathfrak{C}$ is the cycle operator and $\mathfrak{P}$ the set operator.

Translating to exponential generating functions, this gives $$T(z) = \frac{(z^5/5)^{20}}{20!}.$$

The result is now given by (remember that we are using an EGF) $$100! [z^{100}] T(z) = 100! [z^{100}] \frac{z^{100}}{5^{20} 20!} = \frac{100!}{5^{20} 20!},$$ as obtained in the first answer.

I do believe it is useful to employ the symbolic method even in a simple case like this in order to pick up the mechanics and be able to apply it to more complex problems in the future.