AIME Problem 2

I used complimentary counting for this problem. Total number of ways to select the three sleepers from the nine delegates is 9C3 = 84 ways. I found the number of ways that:

1. All of the sleepers are from the same country: There are 0 ways to do this with the people from Mexico (as there are only 2 delegates), 1 way to do it with the people from Canada, and 4 ways to do it with those from the USA . That makes 4+1 = 5 ways.

2. There is one sleeper from each country: This is simply equal to 2 3 4 = 24 ways.

That makes a total of 24+5 = 29 ways to do exactly the opposite of what the problem asks us to solve for. That means that there are 84-29 = 55 ways that exactly two of the three sleepers are from the same country. Therefore, the probability is 55/84 and m+n = 139.

This solution is not mine:

There are three cases:

Case 1: The two Mexican officials fall asleep. Then there are 7 ways to choose from the remaining official for the third sleeper. Case 2: The two Canadian officials fall asleep. There are $\binom{3}{2}$ ways to choose the Canadian officials, then 6 ways to choose the third one for a total of 18 cases. Case 3: The two American officials fall asleep. There are $\binom{4}{2}$ ways to choose the American officials, then 5 ways to choose the third one for a total of 30 cases.

Then, there are $\binom{9}{3}$ ways to choose the three sleepers and the probability is $\dfrac{7+18+30}{\binom{9}{3}} = \frac{55}{84}$ .