Question

$A$ is a set containing elements. $P$ and $Q$ are two subsets of $A$. Then the number of ways of choosing $P$ and $Q$ such that $P \cap Q = \emptyset$ is

• A. 2 2n-2n Cn
• B. 2n
• C. 3n-1
• D. 3n

Answer 1

The Correct Option is D

We are asked to find the number of ways to choose two subsets \( P \) and \( Q \) from a set \( A \) of \( n \) elements, such that \( P \cap Q = \emptyset \), meaning that \( P \) and \( Q \) are disjoint.

Step 1: Understanding the choices for each element of \( A \)
Each element of \( A \) has three possibilities: 1. It can be in \( P \). 2. It can be in \( Q \). 3. It can be in neither \( P \) nor \( Q \). Since \( P \) and \( Q \) are disjoint, no element can be in both \( P \) and \( Q \) simultaneously. 

Step 2: Counting the total number of ways
For each of the \( n \) elements of \( A \), there are 3 choices (either in \( P \), in \( Q \), or in neither). Therefore, the total number of ways to choose \( P \) and \( Q \) is: \[ 3^n \] 

Step 3: Conclusion
Thus, the number of ways to choose \( P \) and \( Q \) such that \( P \cap Q = \emptyset \) is \( 3^n \).

Answer:

\[ \boxed{3^n} \]