Difference between revisions of "2023 AIME II Problems/Problem 11"
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Case 4: <math>N \geq 3</math>. | Case 4: <math>N \geq 3</math>. | ||
− | The number of subsets of <math>\left\{ 1, 2, 3, 4, 5 \right\}</math> that contain at least three elements is <math>\sum_{i=3}^5 \binom{5}{ | + | The number of subsets of <math>\left\{ 1, 2, 3, 4, 5 \right\}</math> that contain at least three elements is <math>\sum_{i=3}^5 \binom{5}{i} = 16</math>. |
Because <math>\mathcal C</math> has 16 elements, we must select all such subsets into <math>\mathcal C</math>. | Because <math>\mathcal C</math> has 16 elements, we must select all such subsets into <math>\mathcal C</math>. | ||
Therefore, the number of solutions in this case is 1. | Therefore, the number of solutions in this case is 1. |
Revision as of 15:44, 5 August 2023
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.