If there must be at least one person in each table, in how many ways can $6$ people be seated around three circular tables?
(We assume tables are indistinguishable)
This problem was an introductory problem to Stirling numbers.
I did not find a solution myself, so looked into the book for it, but I feel it is ambiguous.
It says there are three possible ways to group the number of people (viz. $4 + 1 + 1$, $1 + 2 + 3$ and $2 + 2 + 2$).
Then choosing the people to form groups, and arranging them gives:
$$\tag{QED}
\underbrace{\frac{1}{2}\binom{6}{4}\binom{2}{1}\times3!\times0!\times0!}_{\text{for }4 + 1 + 1} +
\underbrace{\binom{6}{3}\binom{3}{2}\times2!\times1!}_{\text{for }1 + 2 + 3} +
\underbrace{\frac{1}{3!}\binom{6}{2}\binom{4}{2}\times1!\times1!\times1!}_{\text{for }2 + 2 + 2} = 225
$$
Looking at the first term I understood why the $\displaystyle\frac{1}{2}$ is required, and that is because of the following:
\begin{array} & & & & \text{Observer A} & & \\ ^{\text{right}} & & & & & & ^{\text{left}}\\ & & \color{red}{\text{An arrangement of 4}} & \color{green}{\text{1 person}} & \color{purple}{\text{1 person}} & & \\ ^{\text{left}} & & & & & & ^{\text{right}}\\ & & & \text{Observer B} & & & \\ \end{array}
Now, from left to right, Observer A sees the arrangement $\color{purple}{\text{1 person}}$ $\color{green}{\text{1 person}}$ $\color{red}{\text{An arrangement of 4}}$ whereas for Observer B, the same arrangement is $\color{red}{\text{An arrangement of 4}}$ $\color{green}{\text{1 person}}$ $\color{purple}{\text{1 person}}$. As we've counted both of these arrangements, while they are the same, we divide by $2$.
Note that we are giving importance to the arrangement of people on the tables.
Giving the same importance and trying to calculate the last term, we see that we again need to divide by $2$. But it is not what is done. The author decided to divide by $3!$. Now looking at this, we note that
$\color{blue}{\text{2 people }}\color{green}{\text{2 people } } \color{red}{\text{2 people}}$
$\color{green}{\text{2 people }} \color{blue}{\text{2 people }} \color{red}{\text{2 people }}$
$\color{blue}{\text{2 people }} \color{red}{\text{2 people }} \color{green}{\text{2 people }}$
$\color{red}{\text{2 people }} \color{green}{\text{2 people }} \color{blue}{\text{2 people }}$
$\color{red}{\text{2 people }} \color{blue}{\text{2 people }} \color{green}{\text{2 people }}$
$\color{green}{\text{2 people }} \color{red}{\text{2 people }} \color{blue}{\text{2 people }}$
are all counted as the same. This reflects that the arrangements of people on the tables does not matter.
But if this was the case, the $\displaystyle\frac{1}{2}$ in the first term should also be $\displaystyle\frac{1}{3!}$.
What am I missing? Thanks!
The table with four people in the first term is distinct from the tables with one; the two classes of tables cannot be swapped, but the two tables with one person can be. Thus we divide by two in the first term, representing the two ways to permute the two tables with one person.