Find the number of ways of seating n people A1 , A2 , …, An in arbitrary number of circular tables of arbitrary sizes.
Yes, Circular permutations do not matter. For example,
Around a circle A,B,C are three people who sit as ABC, BCA, CAB clockwise are considered equivalent and counted as just one configuration.
And tables are indistinguishable. So, only the configuration of people around the tables matters.
Here is how I counted for n = 4:
4 can be written as,
4 = (4-1)! ways = 6 (4 people around 1 table)
3+1 = (3-1)!*(1-1)! ways = 2 (4 people around 2 tables etc)
2+2 = (2-1)!*(2-1)! ways = 1
2+1+1 = (2-1)!(1-1)!(1-1)! ways = 1
1+1+1+1 = (1-1)!(1-1)!(1-1)!*(1-1)! ways = 1
so total I could count only 11 ways.(Maybe this is flawed in some way? because others think the number of configurations is n! )
Best Answer
Let $F(n)$ be the number of ways of seating $n$ people. We could start by carefully listing the ways, for a few small values of $n$. The listing should be almost explicit. We find that $F(1)=1$, $F(2)=2$, $F(3)=6$, and (perhaps) that $F(4)=24$. Despite the small amount of evidence, the conjecture $F(n)=n!$ is tempting.
We prove the conjecture, in principle by induction. Suppose that we know that for a specific $k$, $F(k)=k!$. Now George, the $(k+1)$-th person, comes along, late as usual. For every possible seating of the $k$ people, we can place George at the table of one of the $k$ people, and immediately to the right of that person ($k$ choices), or we can place George at a table by himself. Thus $$F(k+1)=kF(k)+F(k)=(k+1)F(k)=(k+1)!.$$ Since $F(1)=1$, we conclude that $F(n)=n!$ for all $n$.
Another way: We can use fancier language. For example, note that every permutation of the set $\{1,2,\dots,n\}$ can be expressed uniquely as a product of disjoint cycles. We explicitly include any $1$-cycles. The order in which the product is taken does not matter, since the cycles are disjoint.
Every product of disjoint cycles corresponds to a unique circular seating, and vice-versa. (The people in a cycle determine a table, and their cyclic order determines the order of seating at that table.) This gives an explicit bijection between the permutations of $\{1,2,\dots,n\}$ and the seatings. Thus the number of seatings is equal to $n!$, the number of permutations.