If two sets `A & B` are having `99` elements in common, then the number of elements common to the sets `AxxB` and `BxxA` are

Options

A. `2^(99)`
B. `99^(2)`
C. `100`
D. `18`

Explanation

To solve the problem, we need to find the number of elements 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 intersection of sets \( A \) and \( B \) is denoted as \( A \cap B \).
- We know that \( |A \cap B| = 99 \), which means there are 99 elements that are present in both sets \( A \) and \( B \).

2. **Defining Cartesian Products**:
- The Cartesian product \( A \times B \) consists of all ordered pairs \( (a, b) \) where \( a \in A \) and \( b \in B \).
- Similarly, 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 want to find the intersection \( A \times B \cap B \times A \).
- By the property of Cartesian products, we have:
\[ A \times B \cap B \times A = (A \cap B) \times (A \cap B) \]
- This means that the common elements in \( A \times B \) and \( B \times A \) are formed by taking the elements that are in both \( A \) and \( B \).

4. **Calculating the Number of Common Elements**:
- Since \( |A \cap B| = 99 \), the number of elements in the intersection \( (A \cap B) \times (A \cap B) \) will be:
\[ |(A \cap B) \times (A \cap B)| = |A \cap B| \times |A \cap B| = 99 \times 99 \]
- Therefore, the number of elements common to the sets \( A \times B \) and \( B \times A \) is:
\[ 99^2 = 9801 \]

### Final Answer:
The number of elements common to the sets \( A \times B \) and \( B \times A \) is **9801**.