2023 AIME II Problems/Problem 11
Contents
Problem
Find the number of collections of distinct subsets of with the property that for any two subsets and in the collection,
Solution
Denote by a collection of 16 distinct subsets of . Denote .
Case 1: .
This entails . Hence, for any other set , we have . This is infeasible.
Case 2: .
Let . To get for all . We must have .
The total number of subsets of that contain is . Because contains 16 subsets. We must have . Therefore, for any , we must have . So this is feasible.
Now, we count the number of in this case. We only need to determine . Therefore, the number of solutions is 5.
Case 3: .
Case 3.1: There is exactly one subset in that contains 2 elements.
Denote this subset as . We then put all subsets of that contain at least three elements into , except . This satisfies for any .
Now, we count the number of in this case. We only need to determine . Therefore, the number of solutions is .
Case 3.2: There are exactly two subsets in that contain 2 elements.
They must take the form and .
We then put all subsets of that contain at least three elements into , except and . This satisfies for any .
Now, we count the number of in this case. We only need to determine and . Therefore, the number of solutions is .
Case 3.3: There are exactly three subsets in that contain 2 elements. They take the form , , .
We then put all subsets of that contain at least three elements into , except , , . This satisfies for any .
Now, we count the number of in this case. We only need to determine , , . Therefore, the number of solutions is .
Case 3.4: There are exactly three subsets in that contain 2 elements. They take the form , , .
We then put all subsets of that contain at least three elements into , except , , . This satisfies for any .
Now, we count the number of in this case. We only need to determine , , . Therefore, the number of solutions is .
Case 3.5: There are exactly four subsets in that contain 2 elements. They take the form , , , .
We then put all subsets of that contain at least three elements into , except , , , . This satisfies for any .
Now, we count the number of in this case. We only need to determine , , , . Therefore, the number of solutions is 5.
Putting all subcases together, the number of solutions is this case is .
Case 4: .
The number of subsets of that contain at least three elements is . Because has 16 elements, we must select all such subsets into . Therefore, the number of solutions in this case is 1.
Putting all cases together, the total number of is .
Video Solution
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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