# Difference between revisions of "2011 AIME II Problems/Problem 12"

## Problem 12

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 $m + n$.

## Solution

Use complementary probability and PIE.

If we consider the delegates from each country to be indistinguishable and number the chairs, we have $$\frac{9!}{(3!)^3}$$ total ways to seat the candidates. This comes to: $$\frac{9\cdot8\cdot7\cdot6\cdot5\cdot4}{6\cdot6} = 6\cdot8\cdot7\cdot5 = 30\cdot56.$$

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. This comes to $$\frac{27\cdot6\cdot5\cdot4}6 = 27\cdot 20.$$

Among these there are $$3 \times 9 \times 4$$ ways for candidates from two countries to each sit together. This comes to $$27\cdot 4.$$

Finally, there are $$9 \times 2 = 18.$$ ways for the candidates from all the countries to sit in three blocks (9 clockwise arrangements, and 9 counter-clockwise arrangements).

So, by PIE, the total count of unwanted arrangements is $$27\cdot 20 - 27\cdot 4 + 18 = 16\cdot27 + 18 = 18\cdot25.$$

So the fraction $$\frac mn = \frac{30\cdot 56 - 18\cdot 25}{30\cdot 56} = \frac{56 - 15}{56} = \frac{41}{56}.$$ Thus $m + n = 56 + 41 = \fbox{97.}$