2010 AIME I Problems/Problem 7
Problem
Define an ordered triple of sets to be minimally intersecting if and . For example, is a minimally intersecting triple. Let be the number of minimally intersecting ordered triples of sets for which each set is a subset of . Find the remainder when is divided by .
Note: represents the number of elements in the set .
Solution
Let each pair of two sets have one element in common. Label the common elements as , , . Set will have elements and , set will have and , and set will have and . There are ways to choose values of , and . There are unpicked numbers, and each number can either go in the first set, second set, third set, or none of them. Since we have choices for each of numbers, that gives us .
Finally, , so the answer is .
See also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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All AIME Problems and Solutions |