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I have the following question:

Three Koreans, three New Zealanders and three people from Bosnia are seated at random around a round table. What is the probability that the people in the three groups are seated together?

I have the following:

$(9-1)!/2 = 20160$. This is the total number of ways you can arrange 9 people around a round table.

Then I calculated:

$(9-3)! \cdot 3! \cdot 3 = 12960$

I am considering how to arrange $6$ people around the round table with the (9-3)!, if I get rid of one group of people. The $3!$ comes from how to arrange one group of people, and then you multiply the whole expression by $3$ because there are three groups of people we have to arrange.

To get the probability, I divided the expression by $20160$:

$12960/20160 = 9/14$

However, the answer is $3/280$, and I am not sure why. Any insights are appreciated.

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Suppose Ye-Jin is one of the Koreans. Relative to her, we can seat the remaining eight people in $8!$ ways as we proceed clockwise around the table.

For the favorable cases, observe that if people of each nationality sit together, then the New Zealanders sit either to the left or right of the Koreans. Within each nationality, the three members can be seated in $3!$ ways. Hence, there are $2 \cdot 3!3!3!$ favorable cases.

Thus, the probability that members of each nationality are seated together if people are seated randomly is $$\frac{2 \cdot 3!3!3!}{8!}$$

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