Let `X""=""{1,""2,""3,""4,""5}` . The number of different ordered pairs (Y, Z) that can be formed such that `YsubeX ,""ZsubeX` and `YnnZ` is empty, is (1) `5^2` (2) `3^5` (3) `2^5` (4) `5^3`

Options

A. `5^(2)`
B. `3^(5)`
C. `2^(5)`
D. `5^(3)`

Answer

To solve the problem, we need to find the number of different ordered pairs (Y, Z) such that \( Y \subseteq X \), \( Z \subseteq X \), and \( Y \cap Z = \emptyset \).

Given the set \( X = \{1, 2, 3, 4, 5\} \), we can analyze the possible cases for each element of the set.

### Step-by-Step Solution:

1. **Identify the Elements**:
The set \( X \) has 5 elements: \( 1, 2, 3, 4, 5 \).

2. **Consider Each Element**:
For each element \( a \in X \), we have to decide its membership in the subsets \( Y \) and \( Z \). There are four possible cases for each element:
- Case 1: \( a \) belongs to both \( Y \) and \( Z \) (not allowed since \( Y \cap Z \) must be empty).
- Case 2: \( a \) belongs to \( Y \) but not to \( Z \).
- Case 3: \( a \) does not belong to \( Y \) and does not belong to \( Z \).
- Case 4: \( a \) does not belong to \( Y \) but belongs to \( Z \).

3. **Valid Cases**:
Since Case 1 is not allowed (it violates the condition \( Y \cap Z = \emptyset \)), we are left with three valid cases for each element:
- Case 2: \( a \in Y \) and \( a \notin Z \)
- Case 3: \( a \notin Y \) and \( a \notin Z \)
- Case 4: \( a \notin Y \) and \( a \in Z \)

4. **Count the Choices**:
For each of the 5 elements in \( X \), we have 3 choices (the valid cases).

5. **Calculate the Total Combinations**:
Since the choices for each element are independent, the total number of ordered pairs \( (Y, Z) \) can be calculated as:
\[ \text{Total combinations} = 3^5 \]

6. **Conclusion**:
Thus, the number of different ordered pairs \( (Y, Z) \) such that \( Y \subseteq X \), \( Z \subseteq X \), and \( Y \cap Z = \emptyset \) is \( 3^5 \).

### Final Answer:
The correct option is (2) \( 3^5 \).