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Difference between revisions of "2021 AIME II Problems/Problem 6"

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==Solution==
 
==Solution==
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We can't have a solution without a problem.
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Since
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<math>|A|+|B|-|A \cap B| = |A \cup B|</math>, substituting gives us
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\begin{align*}
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|A| \cdot |B| &= |A \cap B|(|A| + |B| - |A \cap B|)\\
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|A||B| - |A \cap B||A| - |A \cap B||B| + |A \cap B| &= 0\\
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(|A| - |A \cap B|)(|B| - |A \cap B|) &= 0.\\
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\end{align*}.
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Therefore we need <math>A \subseteq B</math> or <math>B \subseteq A</math>. WLOG <math>A\subseteq B</math>, then for each element there are <math>3</math> possibilities, either it is in both <math>A</math> and <math>B</math>, it is in <math>B</math> but not <math>A</math>, or it is in neither <math>A</math> nor <math>B</math>. This gives us <math>3^{5}</math> possibilities, and we multiply by <math>2</math> since it could of also been the other way around. Now we need to subtract the overlaps where <math>A=B</math>, and this case has <math>2^{5}=32</math> ways that could happen. It is <math>32</math> because each number could be in the subset or it could not be in the subset. So the final answer is <math>2\cdot 3^5 - 2^5 = \boxed{454}</math>.
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~ math31415926535
  
 
==See also==
 
==See also==
 
{{AIME box|year=2021|n=II|num-b=5|num-a=7}}
 
{{AIME box|year=2021|n=II|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:24, 22 March 2021

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

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

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