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?


closed as off-topic by Namaste, Leucippus, Claude Leibovici, GNUSupporter 8964民主女神 地下教會, user223391 Dec 23 '17 at 21:59

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The binomial coefficient $\binom63$ counts the number of ways to pick $3$ of the six to sit at a particular table, leaving the other $3$ for the other table. If the tables are indistinguishable, then you might want to divide by $2$, so as not to count $ABC, DEF$ and $DEF,ABC$ as different divisions.

To split six people into three groups of $2$, start with $\binom62$ ways to pick out one group, times $\binom42$ ways to pick the second group, leaving the third group determined. Then, to get rid of redundant choices in different orders, you can divide by $3!$.


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