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?

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#