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

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− | Problem | + | == 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 <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||

− | + | == Solution == | |

− | + | Use complementary probability and PIE. | |

+ | |||

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

+ | |||

+ | Of these there are <cmath> 3 \times 9 \times \frac{6!}{(3!)^2} </cmath> ways to have the candidates of at least some one country sit together. This comes to <cmath>\frac{27\cdot6\cdot5\cdot4}6 = 27\cdot 20.</cmath> | ||

− | + | Among these there are <cmath> 3 \times 9 \times 4 </cmath> ways for candidates from two countries to each sit together. This comes to <cmath> 27\cdot 4. </cmath> | |

+ | |||

+ | Finally, there are <cmath> 9 \times 2 = 18.</cmath> 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 <cmath>27\cdot 20 - 27\cdot 4 + 18 = 16\cdot27 + 18 = 18\cdot25. </cmath> | ||

+ | |||

+ | So the fraction <cmath> \frac mn = \frac{30\cdot 56 - 18\cdot 25}{30\cdot 56} = \frac{56 - 15}{56} = \frac{41}{56}.</cmath> Thus <math>m + n = 56 + 41 = \fbox{97.}</math> |

## Revision as of 01:00, 2 April 2011

## 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 , where and are relatively prime positive integers. Find .

## Solution

Use complementary probability and PIE.

If we consider the delegates from each country to be indistinguishable and number the chairs, we have total ways to seat the candidates. This comes to:

Of these there are ways to have the candidates of at least some one country sit together. This comes to

Among these there are ways for candidates from two countries to each sit together. This comes to

Finally, there are 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

So the fraction Thus