Three Greeks, three Americans and three Italians are seated at random around a round table. What is the probability that the people in the three groups are seated together?

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I know the answer is #3/280# but am unsure how to get to it.

Thanks!

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Answer 1 · 2018-05-26T19:10:47

#3/280#

Let's count the ways all three groups could be seated next to each other, and compare this to the number of ways all 9 could be randomly seated.

We'll number the people 1 through 9, and the groups #A, G, I.#

#stackrel A overbrace(1, 2, 3), stackrel G overbrace(4, 5, 6), stackrel I overbrace(7, 8, 9)#

There are 3 groups, so there are #3! = 6# ways to arrange the groups in a line without disturbing their internal orders:

#AGI, AIG, GAI, GIA, IAG, IGA#

So far this gives us 6 valid permuations.

Within each group, there are 3 members, so there are again #3! = 6# ways to arrange the members within each of the 3 groups:

#123, 132, 213, 231, 312, 321#
#456, 465, 546, 564, 645, 654#
#789, 798, 879, 897, 978, 987#

Combined with the 6 ways to arrange the groups, we now have #6^4# valid permutations so far.

And since we're at a round table, we allow for the 3 arrangements where first group could be "half" on one end and "half" on the other:

#"A A A G G G I I I"#
#"A A G G G I I I A"#
#"A G G G I I I A A"#

The number of total ways to get all 3 groups to be seated together is #6^4 xx 3.#

The number of random ways to arrange all 9 people is #9!#

The probability of randomly choosing one of the "successful" ways is then

#(6xx6xx6xx6xx3)/(9xx8xx7xx6xx5xx4xx3xx2xx1)#

#=(3)/(2xx7xx5xx4)#

#=3/280#