Difference between revisions of "2015 AIME I Problems/Problem 2"
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==Problem== | ==Problem== | ||
The nine delegates to the Economic Cooperation Conference include <math>2</math> officials from Mexico, <math>3</math> officials from Canada, and <math>4</math> 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 <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | The nine delegates to the Economic Cooperation Conference include <math>2</math> officials from Mexico, <math>3</math> officials from Canada, and <math>4</math> 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 <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | The total number of ways to pick <math>3</math> officials from <math>9</math> total is <math>\frac{9!}{(9-3)!} = 9\times8\times7 = 504</math>, 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 <math>7</math> different ways to pick <math>3</math> delegates such that <math>2</math> are from Mexico, simply because there are <math>9-2=7</math> "extra" delegates to choose to be the third sleeper once both from Mexico are sleeping. | ||
+ | * There are <math>3\times6=18</math> ways to pick from Canada, as each Canadian can be left out once and each time one is left out there are <math>9-3=6</math> "extra" people to select one more sleeper 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. | ||
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+ | Thus, the fraction is <math>\frac{7+18+30}{504} = \frac{55}{504}</math> and you don't have to work hard to see if it reduces because the prime factors on top are <math>5</math> and <math>11</math>, neither of which shows up in <math>7</math>, <math>8</math>, or <math>9</math>. Finally, the answer is <math>55+504=\boxed{559}</math>. | ||
== See also == | == See also == |
Revision as of 19:18, 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.
Thus, the fraction is and you don't have to work hard to see if it reduces because the prime factors on top are
and
, neither of which shows up in
,
, or
. Finally, 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.