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?

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

Thanks!

1 Answer
May 26, 2018

3/280

Explanation:

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