Difference between revisions of "2015 AIME I Problems/Problem 2"

(Solution)
m (Solution)
Line 9: Line 9:
 
* 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}{504} = \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>.
  
 
== See also ==
 
== See also ==

Revision as of 18:24, 20 March 2015

Problem

The nine delegates to the Economic Cooperation Conference include $2$ officials from Mexico, $3$ officials from Canada, and $4$ 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 $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

The total number of ways to pick $3$ officials from $9$ total is $\binom{9}{3} = 84$, 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 $7$ different ways to pick $3$ delegates such that $2$ are from Mexico, simply because there are $9-2=7$ "extra" delegates to choose to be the third sleeper once both from Mexico are sleeping.
  • There are $3\times6=18$ ways to pick from Canada, as each Canadian can be left out once and each time one is left out there are $9-3=6$ "extra" people to select one more sleeper from.
  • Lastly, there are $6\times5=30$ ways to choose for the United States. It is easy to count $6$ different ways to pick $2$ of the $4$ Americans, and each time you do there are $9-4=5$ officials left over to choose from.

Thus, the fraction is $\frac{7+18+30}{84} = \frac{55}{84}$. Since this does not reduce, the answer is $55+84=\boxed{139}$.

See also

2015 AIME I (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS