2021 AIME II Problems/Problem 6

Problem

For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . FInd the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy $|A| \cdot |B| = |A \cap B| \cdot |A \cup B|$

Solution

Since $|A|+|B|-|A \cap B| = |A \cup B|$ , substituting gives us \begin{align*} |A| \cdot |B| &= |A \cap B|(|A| + |B| - |A \cap B|)\\ |A||B| - |A \cap B||A| - |A \cap B||B| + |A \cap B| &= 0\\ (|A| - |A \cap B|)(|B| - |A \cap B|) &= 0.\\ \end{align*}. Therefore we need $A \subseteq B$ or $B \subseteq A$ . WLOG $A\subseteq B$ , then for each element there are $3$ possibilities, either it is in both $A$ and $B$ , it is in $B$ but not $A$ , or it is in neither $A$ nor $B$ . This gives us $3^{5}$ possibilities, and we multiply by $2$ since it could of also been the other way around. Now we need to subtract the overlaps where $A=B$ , and this case has $2^{5}=32$ ways that could happen. It is $32$ because each number could be in the subset or it could not be in the subset. So the final answer is $2\cdot 3^5 - 2^5 = \boxed{454}$ .

~ math31415926535

2021 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2021 AIME II Problems/Problem 6

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