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

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==Solution 1==
 
==Solution 1==
−
By PIE, <math>|A|+|B|-|A \cap B| = |A \cup B|</math>, and after some algebra you see that 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>.
+
By PIE, <math>|A|+|B|-|A \cap B| = |A \cup B|</math>. Substituting into the equation and factoring, we get that <math>(|A| - |A \cap B|)(|B| - |A \cap B|) = 0</math>, so therefore <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 have 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>.
  
−
~ math31415926535
+
~math31415926535
  
 
==Solution 2==
 
==Solution 2==
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~ Steven Chen (www.professorchenedu.com)
 
~ Steven Chen (www.professorchenedu.com)
  
−
==Solution 3 (Casework)==
+
==Solution 3 (Principle of Inclusion-Exclusion)==
 
By the <b>Principle of Inclusion-Exclusion (abbreviated as PIE)</b>, we have <math>|A \cup B|=|A|+|B|-|A \cap B|,</math> from which we rewrite the given equation as <cmath>|A| \cdot |B| = |A \cap B| \cdot \left(|A|+|B|-|A \cap B|\right).</cmath>
 
By the <b>Principle of Inclusion-Exclusion (abbreviated as PIE)</b>, we have <math>|A \cup B|=|A|+|B|-|A \cap B|,</math> from which we rewrite the given equation as <cmath>|A| \cdot |B| = |A \cap B| \cdot \left(|A|+|B|-|A \cap B|\right).</cmath>
 
Rearranging and applying Simon's Favorite Factoring Trick give
 
Rearranging and applying Simon's Favorite Factoring Trick give
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Two solutions follow from here:
 
Two solutions follow from here:
  
−
===Solution 3.1 (Combinatorial Identities)===
+
===Solution 3.1 (Binomial Theorem)===
 
The answer is  
 
The answer is  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
\sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] &= 2\underbrace{\sum_{k=0}^{5}\binom{5}{k}2^{5-k}}_{(2+1)^5}-\underbrace{\sum_{k=0}^{5}\binom{5}{k}}_{2^5} \\
+
\sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] &= 2\sum_{k=0}^{5}\binom{5}{k}2^{5-k}-\sum_{k=0}^{5}\binom{5}{k} \\
 +
&=2(2+1)^5-(1+1)^5 \\
 
&=2(243)-32 \\
 
&=2(243)-32 \\
 
&=\boxed{454}.
 
&=\boxed{454}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
 
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
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The answer is  
 
The answer is  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
&\hspace{4.125mm} \left[2\binom{5}{0}2^{5-0}-\binom{5}{0}\right] + \left[2\binom{5}{1}2^{5-1}-\binom{5}{1}\right] + \left[2\binom{5}{2}2^{5-2}-\binom{5}{2}\right] + \left[2\binom{5}{3}2^{5-3}-\binom{5}{3}\right] + \left[2\binom{5}{4}2^{5-4}-\binom{5}{4}\right] + \left[2\binom{5}{5}2^{5-5}-\binom{5}{5}\right] \\
+
&\hspace{5.125mm}\sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] \\
 +
&=\left[2\binom{5}{0}2^{5-0}-\binom{5}{0}\right] + \left[2\binom{5}{1}2^{5-1}-\binom{5}{1}\right] + \left[2\binom{5}{2}2^{5-2}-\binom{5}{2}\right] + \left[2\binom{5}{3}2^{5-3}-\binom{5}{3}\right] + \left[2\binom{5}{4}2^{5-4}-\binom{5}{4}\right] + \left[2\binom{5}{5}2^{5-5}-\binom{5}{5}\right] \\
 
&=\left[2\left(1\right)2^5-1\right] + \left[2\left(5\right)2^4-5\right] + \left[2\left(10\right)2^3-10\right] + \left[2\left(10\right)2^2-10\right] + \left[2\left(5\right)2^1-5\right] + \left[2\left(1\right)2^0-1\right] \\
 
&=\left[2\left(1\right)2^5-1\right] + \left[2\left(5\right)2^4-5\right] + \left[2\left(10\right)2^3-10\right] + \left[2\left(10\right)2^2-10\right] + \left[2\left(5\right)2^1-5\right] + \left[2\left(1\right)2^0-1\right] \\
 
&=63+155+150+70+15+1 \\
 
&=63+155+150+70+15+1 \\
 
&=\boxed{454}.
 
&=\boxed{454}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
~MRENTHUSIASM
 +
 +
==Solution 4 (Simple Bash)==
 +
Proceed with Solution 1 to get <math>(|A| - |A \cap B|)(|B| - |A \cap B|) = 0</math>. WLOG, assume <math>|A| = |A \cap B|</math>. Thus, <math>A \subseteq B</math>.
 +
 +
Since <math>A \subseteq B</math>, if <math>|B| = n</math>, there are <math>2^n</math> possible sets <math>A</math>, and there are also <math>{5 \choose n}</math> ways of choosing such <math>B</math>.
 +
 +
Therefore, the number of possible pairs of sets <math>(A, B)</math> is
 +
 +
<cmath>
 +
\sum_{k=0}^{5} 2^n {5 \choose n}
 +
</cmath>
  
−
~MRENTHUSIASM
+
We can compute this manually since it's only from <math>k=0</math> to <math>5</math>, and computing gives us <math>243</math>. We can double this result for <math>B \subseteq A</math>, and we get <math>2(243) = 486</math>.
 +
 
 +
However, we have double counted the cases where <math>A</math> and <math>B</math> are the same sets. There are <math>32</math> possible such cases, so we subtract <math>32</math> from <math>486</math> to get <math>\boxed{454}</math>.
 +
 
 +
~ adam_zheng
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/jEghPVjyHoc
 +
 
 +
~Interstigation
 +
 
 +
==Video Solution & Set Theory Review==
 +
https://youtu.be/3vcLujj74RM
 +
 
 +
~MathProblemSolvingSkills.com
  
 
==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}}

Latest revision as of 00:02, 29 January 2024

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 1

By PIE, $|A|+|B|-|A \cap B| = |A \cup B|$. Substituting into the equation and factoring, we get that $(|A| - |A \cap B|)(|B| - |A \cap B|) = 0$, so therefore $A \subseteq B$ 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 have 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

Solution 2

We denote $\Omega = \left\{ 1 , 2 , 3 , 4 , 5 \right\}$. We denote $X = A \cap B$, $Y = A \backslash \left( A \cap B \right)$, $Z = B \backslash \left( A \cap B \right)$, $W = \Omega \backslash \left( A \cup B \right)$.

Therefore, $X \cup Y \cup Z \cup W = \Omega$ and the intersection of any two out of sets $X$, $Y$, $Z$, $W$ is an empty set. Therefore, $\left( X , Y , Z , W \right)$ is a partition of $\Omega$.

Following from our definition of $X$, $Y$, $Z$, we have $A \cup B = X \cup Y \cup Z$.

Therefore, the equation

\[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]

can be equivalently written as

\[\left( | X | + | Y | \right) \left( | X | + | Z | \right) = | X | \left( | X | + | Y | + | Z | \right) .\]

This equality can be simplified as

\[| Y | \cdot | Z | = 0 .\]

Therefore, we have the following three cases: (1) $|Y| = 0$ and $|Z| \neq 0$, (2) $|Z| = 0$ and $|Y| \neq 0$, (3) $|Y| = |Z| = 0$. Next, we analyze each of these cases, separately.

Case 1: $|Y| = 0$ and $|Z| \neq 0$.

In this case, to count the number of solutions, we do the complementary counting.

First, we count the number of solutions that satisfy $|Y| = 0$.

Hence, each number in $\Omega$ falls into exactly one out of these three sets: $X$, $Z$, $W$. Following from the rule of product, the number of solutions is $3^5$.

Second, we count the number of solutions that satisfy $|Y| = 0$ and $|Z| = 0$.

Hence, each number in $\Omega$ falls into exactly one out of these two sets: $X$, $W$. Following from the rule of product, the number of solutions is $2^5$.

Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy $|Y| = 0$ minus the number of solutions that satisfy $|Y| = 0$ and $|Z| = 0$, i.e., $3^5 - 2^5$.

Case 2: $|Z| = 0$ and $|Y| \neq 0$.

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., $3^5 - 2^5$.

Case 3: $|Y| = 0$ and $|Z| = 0$.

Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is $2^5$.

By putting all cases together, following from the rule of sum, the total number of solutions is equal to

\begin{align*} \left( 3^5 - 2^5 \right) + \left( 3^5 - 2^5 \right) + 2^5 & = 2 \cdot 3^5 - 2^5 \\ & = \boxed{454} . \end{align*}

~ Steven Chen (www.professorchenedu.com)

Solution 3 (Principle of Inclusion-Exclusion)

By the Principle of Inclusion-Exclusion (abbreviated as PIE), we have $|A \cup B|=|A|+|B|-|A \cap B|,$ from which we rewrite the given equation as \[|A| \cdot |B| = |A \cap B| \cdot \left(|A|+|B|-|A \cap B|\right).\] Rearranging and applying Simon's Favorite Factoring Trick give \begin{align*} |A| \cdot |B| &= |A \cap B|\cdot|A| + |A \cap B|\cdot|B| - |A \cap B|^2 \\ |A| \cdot |B| - |A \cap B|\cdot|A| - |A \cap B|\cdot|B| &= - |A \cap B|^2 \\ \left(|A| - |A \cap B|\right)\cdot\left(|B| - |A \cap B|\right) &=0, \end{align*} from which at least one of the following is true:

  • $|A|=|A \cap B|$
  • $|B|=|A \cap B|$

Let $|A \cap B|=k.$ For each value of $k\in\{0,1,2,3,4,5\},$ we will use PIE to count the ordered pairs $(A,B):$

Suppose $|A|=k.$ There are $\binom{5}{k}$ ways to choose the elements for $A.$ These $k$ elements must also appear in $B.$ Next, there are $2^{5-k}$ ways to add any number of the remaining $5-k$ elements to $B$ (Each element has $2$ options: in $B$ or not in $B.$). There are $\binom{5}{k}2^{5-k}$ ordered pairs for $|A|=k.$ Similarly, there are $\binom{5}{k}2^{5-k}$ ordered pairs for $|B|=k.$

To fix the overcount, we subtract the number of ordered pairs that are counted twice, in which $|A|=|B|=k.$ There are $\binom{5}{k}$ such ordered pairs.

Therefore, there are \[2\binom{5}{k}2^{5-k}-\binom{5}{k}\] ordered pairs for $|A \cap B|=k.$

Two solutions follow from here:

Solution 3.1 (Binomial Theorem)

The answer is \begin{align*} \sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] &= 2\sum_{k=0}^{5}\binom{5}{k}2^{5-k}-\sum_{k=0}^{5}\binom{5}{k} \\ &=2(2+1)^5-(1+1)^5 \\ &=2(243)-32 \\ &=\boxed{454}. \end{align*} ~MRENTHUSIASM

Solution 3.2 (Bash)

The answer is \begin{align*} &\hspace{5.125mm}\sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] \\ &=\left[2\binom{5}{0}2^{5-0}-\binom{5}{0}\right] + \left[2\binom{5}{1}2^{5-1}-\binom{5}{1}\right] + \left[2\binom{5}{2}2^{5-2}-\binom{5}{2}\right] + \left[2\binom{5}{3}2^{5-3}-\binom{5}{3}\right] + \left[2\binom{5}{4}2^{5-4}-\binom{5}{4}\right] + \left[2\binom{5}{5}2^{5-5}-\binom{5}{5}\right] \\ &=\left[2\left(1\right)2^5-1\right] + \left[2\left(5\right)2^4-5\right] + \left[2\left(10\right)2^3-10\right] + \left[2\left(10\right)2^2-10\right] + \left[2\left(5\right)2^1-5\right] + \left[2\left(1\right)2^0-1\right] \\ &=63+155+150+70+15+1 \\ &=\boxed{454}. \end{align*} ~MRENTHUSIASM

Solution 4 (Simple Bash)

Proceed with Solution 1 to get $(|A| - |A \cap B|)(|B| - |A \cap B|) = 0$. WLOG, assume $|A| = |A \cap B|$. Thus, $A \subseteq B$.

Since $A \subseteq B$, if $|B| = n$, there are $2^n$ possible sets $A$, and there are also ${5 \choose n}$ ways of choosing such $B$.

Therefore, the number of possible pairs of sets $(A, B)$ is

\[\sum_{k=0}^{5} 2^n {5 \choose n}\]

We can compute this manually since it's only from $k=0$ to $5$, and computing gives us $243$. We can double this result for $B \subseteq A$, and we get $2(243) = 486$.

However, we have double counted the cases where $A$ and $B$ are the same sets. There are $32$ possible such cases, so we subtract $32$ from $486$ to get $\boxed{454}$.

~ adam_zheng

Video Solution by Interstigation

https://youtu.be/jEghPVjyHoc

~Interstigation

Video Solution & Set Theory Review

https://youtu.be/3vcLujj74RM

~MathProblemSolvingSkills.com

See Also

2021 AIME II (Problems • Answer Key • Resources)
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Problem 7
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