Key Concepts and Formulas

• Principle of Independent Choices (Multiplication Principle): If an event can occur in $m$ ways and another independent event can occur in $n$ ways, then both events can occur in $m \times n$ ways. This extends to multiple independent events.
• Set Theory Basics: Understanding subsets ($\subseteq$) and intersection ($\cap$).
• Condition for Disjoint Sets: Two sets $Y$ and $Z$ are disjoint if their intersection is empty, i.e., $Y \cap Z = \emptyset$.

---

Step-by-Step Solution

Let the given set be $X = \{1, 2, 3, 4, 5\}$. We are asked to find the number of ordered pairs of subsets $(Y, Z)$ such that $Y \subseteq X$, $Z \subseteq X$, and $Y \cap Z = \emptyset$.

Step 1: Analyze the possible placements for each element of $X$. To form the subsets $Y$ and $Z$, we need to decide for each element $x \in X$ where it will be placed. For any element $x$, there are four initial possibilities regarding its membership in $Y$ and $Z$:

1. $x \in Y$ and $x \in Z$.
2. $x \in Y$ and $x \notin Z$.
3. $x \notin Y$ and $x \in Z$.
4. $x \notin Y$ and $x \notin Z$.

Step 2: Apply the constraint $Y \cap Z = \emptyset$. The condition $Y \cap Z = \emptyset$ means that no element can belong to both $Y$ and $Z$ simultaneously. Let's examine the four possibilities from Step 1 in light of this constraint:

1. $x \in Y$ and $x \in Z$: This implies $x \in Y \cap Z$. This violates the condition $Y \cap Z = \emptyset$. So, this placement is not allowed.
2. $x \in Y$ and $x \notin Z$: This implies $x \notin Y \cap Z$. This placement is allowed.
3. $x \notin Y$ and $x \in Z$: This implies $x \notin Y \cap Z$. This placement is allowed.
4. $x \notin Y$ and $x \notin Z$: This implies $x \notin Y \cap Z$. This placement is allowed.

Step 3: Determine the number of valid choices for each element. For each element $x \in X$, there are exactly 3 valid ways it can be placed to satisfy the condition $Y \cap Z = \emptyset$:

• The element belongs exclusively to $Y$.
• The element belongs exclusively to $Z$.
• The element belongs to neither $Y$ nor $Z$.

Step 4: Use the Principle of Independent Choices to find the total number of ordered pairs. The set $X$ has 5 elements. Since the placement of each element is an independent decision, and each element has 3 possible valid placements, we can use the Multiplication Principle. The total number of different ordered pairs $(Y, Z)$ is the product of the number of choices for each of the 5 elements. Total number of pairs = (Choices for element 1) $\times$ (Choices for element 2) $\times$ (Choices for element 3) $\times$ (Choices for element 4) $\times$ (Choices for element 5) $\text{Total number of pairs} = 3 \times 3 \times 3 \times 3 \times 3 = 3^5$

Step 5: Calculate the final value. $3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 9 \times 3 = 81 \times 3 = 243$

The number of different ordered pairs $(Y, Z)$ is $3^5 = 243$.

---

Common Mistakes & Tips

• Confusing ordered pairs with unordered pairs: The question asks for ordered pairs $(Y, Z)$, meaning $(Y, Z)$ is distinct from $(Z, Y)$ unless $Y=Z$. Our method correctly counts ordered pairs by considering the destination of each element for $Y$ and $Z$ separately.
• Forgetting the "empty intersection" condition: If there were no condition on intersection, each element would have 4 choices (in Y only, in Z only, in both, in neither), leading to $4^5$ pairs. The condition $Y \cap Z = \emptyset$ reduces the choices per element from 4 to 3.
• Misinterpreting subset definitions: Ensure you understand that $Y \subseteq X$ means all elements of $Y$ must be in $X$.

---

Summary

The problem requires us to count ordered pairs of subsets $(Y, Z)$ of a set $X$ such that $Y$ and $Z$ are disjoint. For each element in $X$, there are three mutually exclusive and exhaustive possibilities that satisfy the disjoint condition: the element can be in $Y$ only, in $Z$ only, or in neither $Y$ nor $Z$. Since the set $X$ has 5 elements and the choice for each element is independent, by the Principle of Independent Choices, the total number of such ordered pairs is $3$ raised to the power of the number of elements in $X$, which is $3^5$.

The final answer is $\boxed{3^5}$.