If `n` elements are common in sets A and B, then prove that `n^(2)` elements will be common in the set `AxxB and BxxA`.

Explanation

To prove that if `n` elements are common in sets A and B, then `n^2` elements will be common in the sets A × B and B × A, we can follow these steps:

### Step 1: Define the common elements
Let the common elements in sets A and B be represented as:
- \( C = \{c_1, c_2, c_3, \ldots, c_n\} \)
where \( |C| = n \).

### Step 2: Understand the Cartesian products
The Cartesian product of two sets A and B, denoted as \( A \times B \), is defined as:
\[ A \times B = \{(a, b) | a \in A, b \in B\} \]
Similarly, for sets B and A:
\[ B \times A = \{(b, a) | b \in B, a \in A\} \]

### Step 3: Identify the common pairs
Since the elements of C are common to both A and B, we can form pairs from these common elements:
- The pairs in \( A \times B \) that can be formed using elements from C are:
\[ (c_i, c_j) \text{ for } i, j = 1, 2, \ldots, n \]
where both \( c_i \) and \( c_j \) belong to the set C.

### Step 4: Count the common pairs in A × B
The number of such pairs \( (c_i, c_j) \) is:
\[ n \times n = n^2 \]
because for each of the n choices of \( c_i \), there are n choices of \( c_j \).

### Step 5: Identify the common pairs in B × A
Similarly, in the set \( B \times A \), the common pairs formed using elements from C are:
\[ (c_j, c_i) \text{ for } i, j = 1, 2, \ldots, n \]
Again, the number of such pairs is:
\[ n \times n = n^2 \]

### Step 6: Conclude the proof
Thus, the total number of common elements (pairs) in the sets \( A \times B \) and \( B \times A \) is \( n^2 \). Therefore, we have proved that if `n` elements are common in sets A and B, then `n^2` elements will be common in the sets \( A \times B \) and \( B \times A \).