AIME 2015 I, Problem 2

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: One of the best ways to solve this problem is to use PIE, or the Principle of Inclusion and Exclusion. To start off, we know that the denominator, or the total ways to pick $3$ officials, is $\binom{9}{3} = 84$. Now, we find the number of ways that at least $2$ officials are from the same country and then subtract the number of ways all $3$ officials are from the same country. To start with at least $2$ officials, we know: 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 $3\times7=21$ ways to pick from Canada, as we choose $2$ of the $3$ Canadians ($\binom{3}{2} = 3$) and then there are $7$ other options for the third sleeper. Lastly, there are $6\times7=42$ ways to choose for the United States. We can choose two American officials with $\binom{4}{2} = 6$. Then, there are $7$ options for the third sleeper. Now, we want to find the number of ways to have three sleepers from the same country. There are no ways for the $3$ sleepers to be from Mexico because there are only $2$ Mexican officials. Hence, we get $0$ ways. There is only $1$ way to pick all $3$ from Canada because there are exactly $3$ Canadian officials. We now consider the number of times we originally counted this, which after inspection, is $3$, so we have $1 * 3 = 3$. Lastly, there are $4$ ways to choose all $3$ officials from the United States ($\binom{4}{3} = 4$). Once again, we counted this $3$ times, so we have $4*3 = 12$. Thus, the fraction is $\frac{7+21+42-0-3-12}{84} = \frac{55}{84}$, and our answer is $55+84=\fbox{139}$. Note: Similar to Solution 2, we could also have grouped each of the countries in terms of the ways we have for at least $2$ officials to sleep and for all $3$ to sleep. Mexico would have $7 - 0 = 7$ ways for exactly 2 officials to be there, Canada with $21 - 3 = 18$ ways, and the United States with $42 - 12 = 30$ ways. Solution by: armang32324

Answer: 139