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. **Identify the Given Information**:
- We know that the number of elements in the intersection of sets \( A \) and \( B \) is given as:
\[ N(A \cap B) = 99 \]

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

3. **Find the Intersection of the Cartesian Products**:
- We want to find the intersection \( A \times B \cap B \times A \).
- Using the property of Cartesian products, we can express this as:
\[ A \times B \cap B \times A = (A \cap B) \times (A \cap B) \]
- This means we are looking for pairs \( (x, y) \) such that both \( x \) and \( y \) are from the intersection \( A \cap B \).

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

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