1998 AIME Problems/Problem 4

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Problem

Nine tiles are numbered respectively. Each of three players randomly selects and keeps three of the tiles, and sums those three values. The probability that all three players obtain an odd sum is where and are relatively prime positive integers. Find

Solution

There are 5 odd tiles and 4 even tiles. In order for a player to have an odd sum, she must have an odd number of odd tiles. Thus, the only possibility is that one player gets three odd tiles and the other two players get two even and one odd tile. We count the number of ways this can happen:

We have 3 choices for the player who can get all odd tiles, and choices for the tiles that he gets. The other two odd tiles can be distributed to the other two players in 2 ways, and the even tiles can be distributed between them in ways. This gives us a total of possibilities.

In order to calculate the probability, we need to know the total number of possible distributions for the tiles. The first player needs three tiles which we can give him in ways, and the second player needs three of the remaining six, which we can give him in ways.

Thus, the total probability is so the answer is .

See also