# 2015 AIME I Problems/Problem 2

## 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 $55+84=\boxed{139}$.

## Solution 2

The total number of ways to pick $3$ officials from $9$ total is $84$. 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 $1$, $2$, or $3$ countries.

• If the sleeping officials are from a single country, they can be from Canada in $1$ way and from the United States in $4$ ways, so there are $5$ total possibilities.
• If the sleeping officials are from $3$ different countries, there must be one from each. So there are $2 \cdot 3 \cdot 4 = 24$ total possibilities.

Out of $84$ total, $5+24=29$ possibilities are different from the case we are looking for, so there are $84-29=55$ total ways to choose 22 officials from one country and a single official from another country. As $55$ and $84$ share no common factors, we have $55+84=\boxed{139}$.

## Solution 3

The first thing we need to do is to figure out the total number of ways we can pick the 3 delegates that fell asleep. We can do this using combinations: $\dbinom{9}{3} = 84$

## See also

 2015 AIME I (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 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.

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