Difference between revisions of "2015 AIME I Problems/Problem 2"
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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>. | 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>. | ||
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| + | ==Solution 2== | ||
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| + | The total number of ways to pick <math>3</math> officials from <math>9</math> total is <math>84</math>. We note that two sleepers are asleep and one is awake if and only if the sleepers come from two distinct countries. The sleeping officials can be from either <math>1</math>, <math>2</math>, or <math>3</math> countries. | ||
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| + | * If the sleeping officials are from a single country, they can be from Canada in <math>1</math> way and from the United States in <math>4</math> ways, so there are <math>5</math> total possibilities. | ||
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| + | * If the sleeping officials are from <math>3</math> different countries, there must be one from each. So there are <math>2 \cdot 3 \cdot 4 = 24</math> total possibilities. | ||
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| + | Out of <math>84</math> total, <math>5+24=29</math> possibilities are different from the case we are looking for, so there are <math>84-29=55</math> total ways to choose 22 officials from one country and a single official from another country. As <math>55</math> and <math>84</math> share no common factors, we have <math>55+84=\boxed{139}</math>. | ||
== See also == | == See also == | ||
Revision as of 19:49, 21 March 2015
Contents
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 
.
Solution 2
The total number of ways to pick officials from total is 
. We note that two sleepers are asleep and one is awake if and only if the sleepers come from two distinct countries. The sleeping officials can be from either 
, , or countries.
- If the sleeping officials are from a single country, they can be from Canada in
way and from the United States in ways, so there are 
total possibilities.
total possibilities.- If the sleeping officials are from different countries, there must be one from each. So there are
total possibilities.
total possibilities.Out of total, 
possibilities are different from the case we are looking for, so there are 
total ways to choose 22 officials from one country and a single official from another country. As 
and share no common factors, we have .
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. 
