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.