Question

Nine tiles are numbered $1, 2, 3, \ldots, 9,$ 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 $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

Solution:

A player's three tiles have an odd sum exactly when the player holds an odd number of odd tiles — one or three. The nine tiles include five odd and four even, and the only way to split five odd tiles into three groups of size one or three is $3 + 1 + 1.$

Count favorable deals: choose which player gets three odd tiles ( $3$ ways), choose that player's odd tiles ( $\binom{5}{3} = 10$ ways), give one of the two remaining odd tiles to each other player ( $2$ ways), then split the four even tiles two and two between those players ( $\binom{4}{2} = 6$ ways), for $3 \cdot 10 \cdot 2 \cdot 6 = 360$ deals. The total number of deals is $\binom{9}{3}\binom{6}{3} = 84 \cdot 20 = 1680.$

The probability is $\frac{360}{1680} = \frac{3}{14},$ so $m + n = 3 + 14 = 17.$

1998 AIME Problem 4

Solution:

Problem 4 in Other Years