2023 AIME II Problems/Problem 11

Problem

Find the number of collections of $16$ distinct subsets of $\{1,2,3,4,5\}$ with the property that for any two subsets $X$ and $Y$ in the collection, $X \cap Y \not= \emptyset.$

Solution

Denote by $\mathcal C$ a collection of 16 distinct subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ . Denote $N = \min \left\{ |S|: S \in \mathcal C \right\}$ .

Case 1: $N = 0$ .

This entails $\emptyset \in \mathcal C$ . Hence, for any other set $A \in \mathcal C$ , we have $\emptyset \cap A = \emptyset$ . This is infeasible.

Case 2: $N = 1$ .

Let $\{a_1\} \in \mathcal C$ . To get $\{a_1\} \cap A \neq \emptyset$ for all $A \in \mathcal C$ . We must have $a_1 \in \mathcal A$ .

The total number of subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain $a_1$ is $2^4 = 16$ . Because $\mathcal C$ contains 16 subsets. We must have $\mathcal C = \left\{ \{a_1\} \cup A : \forall \ A \subseteq \left\{ 1, 2, 3, 4, 5 \right\} \backslash \left\{a_1 \right\} \right\}$ . Therefore, for any $X, Y \in \mathcal C$ , we must have $X \cap Y \supseteq \{a_1\}$ . So this is feasible.

Now, we count the number of $\mathcal C$ in this case. We only need to determine $a_1$ . Therefore, the number of solutions is 5.

Case 3: $N = 2$ .

Case 3.1: There is exactly one subset in $\mathcal C$ that contains 2 elements.

Denote this subset as $\left\{ a_1, a_2 \right\}$ . We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ . This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$ .

Now, we count the number of $\mathcal C$ in this case. We only need to determine $\left\{ a_1, a_2 \right\}$ . Therefore, the number of solutions is $\binom{5}{2} = 10$ .

Case 3.2: There are exactly two subsets in $\mathcal C$ that contain 2 elements.

They must take the form $\left\{ a_1, a_2 \right\}$ and $\left\{ a_1, a_3 \right\}$ .

We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ and $\left\{ a_2, a_4, a_5 \right\}$ . This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$ .

Now, we count the number of $\mathcal C$ in this case. We only need to determine $\left\{ a_1, a_2 \right\}$ and $\left\{ a_1, a_3 \right\}$ . Therefore, the number of solutions is $5 \cdot \binom{4}{2} = 30$ .

Case 3.3: There are exactly three subsets in $\mathcal C$ that contain 2 elements. They take the form $\left\{ a_1, a_2 \right\}$ , $\left\{ a_1, a_3 \right\}$ , $\left\{ a_1, a_4 \right\}$ .

We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ , $\left\{ a_2, a_4, a_5 \right\}$ , $\left\{ a_2, a_3, a_5 \right\}$ . This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$ .

Now, we count the number of $\mathcal C$ in this case. We only need to determine $\left\{ a_1, a_2 \right\}$ , $\left\{ a_1, a_3 \right\}$ , $\left\{ a_1, a_4 \right\}$ . Therefore, the number of solutions is $5 \cdot \binom{4}{3} = 20$ .

Case 3.4: There are exactly three subsets in $\mathcal C$ that contain 2 elements. They take the form $\left\{ a_1, a_2 \right\}$ , $\left\{ a_1, a_3 \right\}$ , $\left\{ a_2, a_3 \right\}$ .

We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ , $\left\{ a_2, a_4, a_5 \right\}$ , $\left\{ a_1, a_4, a_5 \right\}$ . This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$ .

Now, we count the number of $\mathcal C$ in this case. We only need to determine $\left\{ a_1, a_2 \right\}$ , $\left\{ a_1, a_3 \right\}$ , $\left\{ a_2, a_3 \right\}$ . Therefore, the number of solutions is $\binom{5}{3} = 10$ .

Case 3.5: There are exactly four subsets in $\mathcal C$ that contain 2 elements. They take the form $\left\{ a_1, a_2 \right\}$ , $\left\{ a_1, a_3 \right\}$ , $\left\{ a_1, a_4 \right\}$ , $\left\{ a_1, a_5 \right\}$ .

We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ , $\left\{ a_2, a_4, a_5 \right\}$ , $\left\{ a_1, a_4, a_5 \right\}$ , $\left\{ a_2, a_3, a_4 \right\}$ . This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$ .

Now, we count the number of $\mathcal C$ in this case. We only need to determine $\left\{ a_1, a_2 \right\}$ , $\left\{ a_1, a_3 \right\}$ , $\left\{ a_1, a_4 \right\}$ , $\left\{ a_1, a_5 \right\}$ . Therefore, the number of solutions is 5.

Putting all subcases together, the number of solutions is this case is $10 + 30 + 20 + 10 + 5 = 75$ .

Case 4: $N \geq 3$ .

The number of subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements is $\sum_{i=3}^5 \binom{5}{3} = 16$ . Because $\mathcal C$ has 16 elements, we must select all such subsets into $\mathcal C$ . Therefore, the number of solutions in this case is 1.

Putting all cases together, the total number of $\mathcal C$ is $5 + 75 + 1 = \boxed{\textbf{(081) }}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/FH2QUGVNchw

~MathProblemSolvingSkills.com

2023 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2023 AIME II Problems/Problem 11

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