Problem: If at least one person must be on each table, what is the number of ways to sit six people around $(i)$ two tables $(ii)$ three tables? (it is assume that the tables are indistinguishable)
Here's a solution.
for $(i)$, we consider 3 cases
(1) 5 + 1, (2), 4 + 2 and (3) 3 + 3
around each table, the number of ways for (1) is $6 \choose 5$, for (2) is $6 \choose 4$, but what about (3)?
What is the number of ways to divide 6 people into 2 groups of size three each?(answer with explanations please).
Also, for $(ii)$, how do we handle the case of sitting 2 + 2 + 2 persons around 3 tables?