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

(→‎Solution 2)
(→‎Solution 2)
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Next, we analyze each of these cases, separately.
 
Next, we analyze each of these cases, separately.
  
−
Case 1: <math>|Y| = 0</math> and <math>|Z| \neq 0</math>.
+
\bf{Case 1}: <math>|Y| = 0</math> and <math>|Z| \neq 0</math>.
  
 
In this case, to count the number of solutions, we do the complementary counting.  
 
In this case, to count the number of solutions, we do the complementary counting.  
Line 55: Line 55:
 
Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy <math>|Y| = 0</math> minus the number of solutions that satisfy <math>|Y| = 0</math> and <math>|Z| = 0</math>, i.e., <math>3^5 - 2^5</math>.
 
Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy <math>|Y| = 0</math> minus the number of solutions that satisfy <math>|Y| = 0</math> and <math>|Z| = 0</math>, i.e., <math>3^5 - 2^5</math>.
  
−
Case 2: <math>|Z| = 0</math> and <math>|Y| \neq 0</math>.  
+
\bf{Case 2}: <math>|Z| = 0</math> and <math>|Y| \neq 0</math>.  
  
 
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., <math>3^5 - 2^5</math>.
 
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., <math>3^5 - 2^5</math>.
  
−
Case 3: <math>|Y| = 0</math> and <math>|Z| = 0</math>.
+
\bf{Case 3}: <math>|Y| = 0</math> and <math>|Z| = 0</math>.
  
 
Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is <math>2^5</math>.
 
Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is <math>2^5</math>.

Revision as of 20:31, 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 1

By PIE, $|A|+|B|-|A \cap B| = |A \cup B|$, and after some algebra you see that 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

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.

\bf{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., $3^5 - 2^5$.

\bf{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., $3^5 - 2^5$.

\bf{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)

2021 AIME II (Problems • Answer Key • Resources)
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