Question

A hotel packed a breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls, and, once they were wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability that each guest got one roll of each type is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

答案

Solution:

Fill the first guest's bag one roll at a time. The first roll can be anything; the second must avoid the $2$ remaining rolls of the first roll's type, succeeding with probability $\frac{6}{8};$ and the third must be one of the $3$ rolls of the missing type among the remaining $7.$ So the first bag has one roll of each type with probability $\frac{6}{8} \cdot \frac{3}{7} = \frac{9}{28}.$

Given that, six rolls remain, two of each type, and the same argument gives $\frac{4}{5} \cdot \frac{2}{4} = \frac{2}{5}$ for the second bag. The third bag is then automatically one of each type. The probability is $\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70},$ so $m + n = 9 + 70 = 79.$

2005 AIME II Problem 2

Solution:

Problem 2 in Other Years