If two sets A and B are having 99 elements in common, then the number of elements common to each of the sets `A xx B` and `B xx A` are

Options

A. A) `2^(99)`
B. B) `99^2`
C. C) 100
D. D) 19

Answer

To solve the problem, we need to find the number of elements that are common to the sets \( A \times B \) and \( B \times A \), given that the two sets \( A \) and \( B \) have 99 elements in common.

### Step-by-Step Solution:

1. **Understanding the Sets**:
- Let \( A \) and \( B \) be two sets.
- The number of elements in common between \( A \) and \( B \) is given as \( |A \cap B| = 99 \).

2. **Defining the Cartesian Products**:
- The Cartesian product \( A \times B \) consists of all ordered pairs \( (a, b) \) where \( a \in A \) and \( b \in B \).
- The Cartesian product \( B \times A \) consists of all ordered pairs \( (b, a) \) where \( b \in B \) and \( a \in A \).

3. **Finding the Intersection**:
- We need to find the intersection \( (A \times B) \cap (B \times A) \).
- An element in \( (A \times B) \) is of the form \( (a, b) \) and an element in \( (B \times A) \) is of the form \( (b, a) \).
- For \( (a, b) \) to be in both sets, \( a \) must be in \( B \) and \( b \) must be in \( A \).

4. **Identifying Common Elements**:
- Since \( a \) and \( b \) are common elements from the intersection \( A \cap B \), we can denote these common elements as \( x_1, x_2, \ldots, x_{99} \).
- Each of these 99 elements can form pairs in both \( A \times B \) and \( B \times A \).

5. **Calculating the Number of Common Pairs**:
- The number of pairs \( (x_i, x_j) \) where both \( x_i \) and \( x_j \) are from the intersection \( A \cap B \) is \( 99 \times 99 \) because:
- There are 99 choices for \( a \) (from \( A \cap B \)) and 99 choices for \( b \) (from \( A \cap B \)).
- Therefore, the number of common elements in \( A \times B \) and \( B \times A \) is \( 99^2 \).

6. **Final Answer**:
- The number of elements common to \( A \times B \) and \( B \times A \) is \( 99^2 = 9801 \).