# 2011 AIME Problem 12 — Probability: 9 delegates around a round table

Nine delegates, three each from three different countries, randomly
select chairs at a round table that seats nine people. Let the
probability that each delegate sits next to at least one delegate from
another country be $\frac{m}{n}$, where m and n are relatively prime
positive integers. Find $m + n$.

I was going through the solution at AoPS, and I didn’t understand several things.

We have $$\frac{9!}{(3!)^3} = \frac{9\cdot8\cdot7\cdot6\cdot5\cdot4}{6\cdot6} = 6\cdot8\cdot7\cdot5 = 30\cdot56$$ total ways to seat the candidates.

Because the candidates are seated in a circle, shouldn’t it be $$\frac{\frac{9!}{(3!)^3}}{9}$$ to account for the rotations?

Of these, there are $$3 \times 9 \times \frac{6!}{(3!)^2}$$ ways to have the candidates of at least some one country sit together.

My question comes from why they used $$\frac{9\cdot 6!}{3!3!}$$

I believe their reasoning is that there are nine ways to choose the three consecutive chairs that the three delegates from a specific country A sit in, and then $\binom{6}{3}$ ways to seat the other country and then the last country sits in the open spots.

However, I tried to do this a different way, by considering the three delegates from country A as one entity, and then you have the 6 seats + 1 group to arrange around the circle, which can be done in $\frac{7!}{3!3!}$ ways.

However, this does not match with their expression. Why?

#### Solutions Collecting From Web of "2011 AIME Problem 12 — Probability: 9 delegates around a round table"

They are making two simplifying assumptions in their solution, that the seats are numbered and that people from the same country are identical.

1) They didn’t divide by $9$ in finding their total because they were assuming the seats were numbered; otherwise they would have.

2) Your expression of $\frac{7!}{3!3!}$ would work fine if the people were arranged in a straight line. Notice that in this case the 3 people sitting together have 7 choices for their positions: $123, 234, \cdots, 789$.
In a circle, though, they have 9 choices instead: $123, 234, \cdots, 789, 891, 912$.

If we assumed that the 9 people were all distinct and we didn’t number the seats, then we would get the same probability of

$\displaystyle 1-\frac{\binom{3}{1}3!6!-\binom{3}{2}4(3!)^{3}+2(3!)^{3}}{8!}=1-\frac{10800}{40320}=1-\frac{15}{56}=\frac{41}{56}$.