Difference between revisions of "2015 AIME I Problems/Problem 2"
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* Lastly, there are <math>6\times5=30</math> ways to choose for the United States. It is easy to count <math>6</math> different ways to pick <math>2</math> of the <math>4</math> Americans, and each time you do there are <math>9-4=5</math> officials left over to choose from. | * Lastly, there are <math>6\times5=30</math> ways to choose for the United States. It is easy to count <math>6</math> different ways to pick <math>2</math> of the <math>4</math> Americans, and each time you do there are <math>9-4=5</math> officials left over to choose from. | ||
| − | Thus, the fraction is <math>\frac{7+18+30}{ | + | Thus, the fraction is <math>\frac{7+18+30}{84} = \frac{55}{84}</math>. Since this does not reduce, the answer is <math>55+84=\boxed{139}</math>. |
== See also == | == See also == | ||
Revision as of 18:24, 20 March 2015
Problem
The nine delegates to the Economic Cooperation Conference include 
officials from Mexico, 
officials from Canada, and 
officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
The total number of ways to pick officials from 
total is 
, which will be our denominator. Now we can consider the number of ways for exactly two sleepers to be from the same country for each country individually and add them to find our numerator:
- There are
different ways to pick 
delegates such that 
are from Mexico, simply because there are 
"extra" delegates to choose to be the third sleeper once both from Mexico are sleeping. - There are
ways to pick from Canada, as each Canadian can be left out once and each time one is left out there are 
"extra" people to select one more sleeper from. - Lastly, there are
ways to choose for the United States. It is easy to count 
different ways to pick of the 
Americans, and each time you do there are 
officials left over to choose from.
different ways to pick 
"extra" delegates to choose to be the third sleeper once both from Mexico are sleeping.
ways to pick from Canada, as each Canadian can be left out once and each time one is left out there are 
"extra" people to select one more sleeper from.
ways to choose for the United States. It is easy to count 
different ways to pick 
officials left over to choose from.Thus, the fraction is 
. Since this does not reduce, the answer is 
.
See also
| 2015 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
