2021 AIME II Problems/Problem 6
Contents
Problem
For any finite set , let denote the number of elements in . Find the number of ordered pairs such that and are (not necessarily distinct) subsets of that satisfy
Solution 1
By PIE, , and after some algebra you see that we need or . WLOG , then for each element there are possibilities, either it is in both and , it is in but not , or it is in neither nor . This gives us possibilities, and we multiply by since it could have also been the other way around. Now we need to subtract the overlaps where , and this case has ways that could happen. It is because each number could be in the subset or it could not be in the subset. So the final answer is .
~math31415926535
Solution 2
We denote . We denote , , , .
Therefore, and the intersection of any two out of sets , , , is an empty set. Therefore, is a partition of .
Following from our definition of , , , we have .
Therefore, the equation
can be equivalently written as
This equality can be simplified as
Therefore, we have the following three cases: (1) and , (2) and , (3) . Next, we analyze each of these cases, separately.
Case 1: and .
In this case, to count the number of solutions, we do the complementary counting.
First, we count the number of solutions that satisfy .
Hence, each number in falls into exactly one out of these three sets: , , . Following from the rule of product, the number of solutions is .
Second, we count the number of solutions that satisfy and .
Hence, each number in falls into exactly one out of these two sets: , . Following from the rule of product, the number of solutions is .
Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy minus the number of solutions that satisfy and , i.e., .
Case 2: and .
This case is symmetric to Case 1. Therefore, the number of solutions in this case is the same as the number of solutions in Case 1, i.e., .
Case 3: and .
Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is .
By putting all cases together, following from the rule of sum, the total number of solutions is equal to
~ Steven Chen (www.professorchenedu.com)
Solution 3 (Principle of Inclusion-Exclusion)
By the Principle of Inclusion-Exclusion (abbreviated as PIE), we have from which we rewrite the given equation as Rearranging and applying Simon's Favorite Factoring Trick give from which at least one of the following is true:
Let For each value of we will use PIE to count the ordered pairs
Suppose There are ways to choose the elements for These elements must also appear in Next, there are ways to add any number of the remaining elements to (Each element has options: in or not in ). There are ordered pairs for Similarly, there are ordered pairs for
To fix the overcount, we subtract the number of ordered pairs that are counted twice, in which There are such ordered pairs.
Therefore, there are ordered pairs for
Two solutions follow from here:
Solution 3.1 (Binomial Theorem)
The answer is ~MRENTHUSIASM
Solution 3.2 (Bash)
The answer is ~MRENTHUSIASM
Video Solution by Interstigation
~Interstigation
Video Solution & Set Theory Review
~MathProblemSolvingSkills.com
See Also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.