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 $\sum_{i=3}^5 \binom{5}{i} = 16$ . 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 $5 + 75 + 1 = \boxed{\textbf{(081) }}$ .

Solution 2

Denote the $A$ as $\{ 1,2,3,4,5 \}$ and the collection of subsets as $S$ .

Case 1: There are only sets of size $3$ or higher in $S$ : Any two sets in $S$ must have at least one element common to both of them (since $3+3>5$ ). Since there are $16$ subsets of $A$ that have size $3$ or higher, there is only one possibility for this case.

Case 2: There are only sets of size $2$ or higher in $S$ : Firstly, there cannot be a no size $2$ set $S$ , or it will be overcounting the first case.

If there is only one such size $2$ set, there are $10$ ways to choose it. That size $2$ set, say $X$ , cannot be in $S$ with $Y = A/X$ (a three element set). Thus, there are only $15$ possible size $3$ subsets that can be in $S$ , giving us $10$ for this case.

If there are two sets with size $2$ in $S$ , we can choose the common elements of these two subsets in $5$ ways, giving us a total of $5\cdot 6 = 30$ .

If there are three sets with size $2$ , they can either share one common element, which can be done in $5 \cdot 4 = 20$ ways, or they can share pairwise common elements (sort of like a cycle), which can be done $\binom{5}{2} = 10$ ways. In total, we have $30$ possibilities.

If there are four sets with size $2$ , they all have to share one common element, which can be done in $5\cdot 1$ ways.

Thus, summing everything up, this will give us $75$ possible sets $S$

Case 3: There is a set with size $1$ in $S$ : Notice that be at most one size $1$ subset. There are $5$ ways to choose that single element set. Say it's $\{ 1\}$ . All other subsets in $S$ must have a $1$ in them, but there are only $2^4-1 = 15$ of them. Thus this case yields $5\cdot 1 = 5$ possibilities.

Thus, the total number of sets $S$ would be $1+75+5 = \boxed{\textbf{(081)}}$ .

~sml1809

Video Solution

https://youtu.be/FH2QUGVNchw

2023 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2023 AIME II Problems/Problem 11

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Solution

Solution 2

Video Solution

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