Difference between revisions of "2023 AIME II Problems/Problem 11"
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Denote the <math>A</math> as <math>\{ 1,2,3,4,5 \}</math> and the collection of subsets as <math>S</math>. | Denote the <math>A</math> as <math>\{ 1,2,3,4,5 \}</math> and the collection of subsets as <math>S</math>. | ||
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<b>Case 1: There are only sets of size <math>3</math> or higher in <math>S</math>: </b> | <b>Case 1: There are only sets of size <math>3</math> or higher in <math>S</math>: </b> | ||
Any two sets in <math>S</math> must have at least one element common to both of them (since <math>3+3>5</math>). Since there are <math>16</math> subsets of <math>A</math> that have size <math>3</math> or higher, there is only one possibility for this case. | Any two sets in <math>S</math> must have at least one element common to both of them (since <math>3+3>5</math>). Since there are <math>16</math> subsets of <math>A</math> that have size <math>3</math> or higher, there is only one possibility for this case. | ||
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<b>Case 2: There are only sets of size <math>2</math> or higher in <math>S</math>: </b> | <b>Case 2: There are only sets of size <math>2</math> or higher in <math>S</math>: </b> | ||
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Thus, summing everything up, this will give us <math>75</math> possible sets <math>S</math> | Thus, summing everything up, this will give us <math>75</math> possible sets <math>S</math> | ||
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<b>Case 3: There is a set with size <math>1</math> in <math>S</math>: </b> | <b>Case 3: There is a set with size <math>1</math> in <math>S</math>: </b> |
Revision as of 16:58, 6 January 2024
Contents
[hide]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 .
Solution 2
Denote the as and the collection of subsets as .
Case 1: There are only sets of size or higher in :
Any two sets in must have at least one element common to both of them (since ). Since there are subsets of that have size or higher, there is only one possibility for this case.
Case 2: There are only sets of size or higher in :
Firstly, there cannot be a no size set , or it will be overcounting the first case.
If there is only one such size set, there are ways to choose it. That size set, say , cannot be in with (a three element set). Thus, there are only possible size subsets that can be in , giving us for this case.
If there are two sets with size in , we can choose the common elements of these two subsets in ways, giving us a total of .
If there are three sets with size , they can either share one common element, which can be done in ways, or they can share pairwise common elements (sort of like a cycle), which can be done ways. In total, we have possibilities.
If there are four sets with size , they all have to share one common element, which can be done in ways.
Thus, summing everything up, this will give us possible sets
Case 3: There is a set with size in :
Notice that be at most one size subset. There are ways to choose that single element set. Say it's . All other subsets in must have a in them, but there are only of them. Thus this case yields possibilities.
Thus, the total number of sets would be .
~sml1809
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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