2015 AIME I Problem 2

We first count the total numer $T$ of combinations

$T=\binom{9}{3}=84$

And the number $F$ of favorable cases, counting the number of ways we can choose $2$ delegates from a country and multiplying it by the number of ways we can choose the third one from another country

Mexico: $F_M=\binom{2}{2}\cdot(9-2)=7$

Canada: $F_C=\binom{3}{2}\cdot(9-3)=18$

United States: $F_{US}=\binom{4}{2}\cdot(9-4)=30$

$\Longrightarrow F=7+18+30=55$

Therefore the probability is

$P=\dfrac{F}{T}=\dfrac{55}{84}=\dfrac{m}{n}$

$\Longrightarrow m+n=55+84=\boxed{139}$

We count the number of ways that there are NOT exactly two sleepers from the same country.

The first way is to have 1 sleeper from each country. There are 2 people to choose from for MX, 3 from Canada, and 4 from US, so there are 2 3 4=24 ways to do this.

The last way is to have 3 sleepers from 1 country. This is not possible for MX, there's only 1 way to do this for Canada, and there are 4 ways to do this for US (there are four people total and we need to pick three, so there are four choices for who to leave out).

Now, let's count up the total number of ways to pick sleepers. We have to choose 3 people from a total of 9, which works out to 9!/(3!6!) = 84.

Lastly, there were 29 ways to not have exactly two people from one country, and 84 possibilities total, so there are 55 ways to have exactly two people from one country. So, the probability is 55/84, and 55+84=139.