Question

At a particular receptionist, 9 guests are to stand at 3 identical round cocktail tables. How many ways can this be done if there must be at least 2 people at each table?

Answer 1

Answer on Question #45807 – Math – Combinatorics | Number Theory

Question.

At a particular receptionist, 9 guests are to stand at 3 identical round cocktail tables. How many ways can this be done if there must be at least 2 people at each table?

Solution.

We choose three of the nine guests. We can do this in $C_9^3$ ways. Now, make them sit at three tables so that each table had one guest. Since the tables are identical we can do it in $\frac{3 \cdot 2 \cdot 1}{3!} = 1$ ways. Next we choose three of six guests. We can do this in $C_6^3$ ways. Make them again one at each of the three tables. Now the order is important so we can do it in $3 \cdot 2 \cdot 1 = 6$ ways. We can accommodate arbitrarily the remaining three guests in $3 \cdot 3 \cdot 3 = 27$ ways. We have by the rule of multiplication $C_9^3 \cdot 1 \cdot C_6^3 \cdot 6 \cdot 27 = \frac{9!}{3!6!} \cdot \frac{6!}{3!3!} \cdot 6 \cdot 27 = 272160$ ways.

Answer: 272160 ways.

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