# 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 $\frac{3}{280}$ but am unsure how to get to it. Thanks!

May 26, 2018

$\frac{3}{280}$

#### 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:

$A G I , A I G , G A I , G I A , I A G , I G A$

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:

$\text{A A A G G G I I I}$
$\text{A A G G G I I I A}$
$\text{A G G G I I I A A}$

The number of total ways to get all 3 groups to be seated together is ${6}^{4} \times 3.$

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

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

$\frac{6 \times 6 \times 6 \times 6 \times 3}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$

$= \frac{3}{2 \times 7 \times 5 \times 4}$

$= \frac{3}{280}$