A and B are two sets and `n(AxxB)` =6 . If there elements of `AxxB` are (1,2) ,(2,3),(3,3), then find `AxxB` and `BxxA`.

Explanation

To solve the problem step by step, we will find the sets A and B based on the given Cartesian product \( A \times B \) and then compute both \( A \times B \) and \( B \times A \).

### Step 1: Understand the Cartesian Product
The Cartesian product \( A \times B \) consists of ordered pairs where the first element comes from set A and the second element comes from set B. The number of elements in \( A \times B \) is given by the formula:
\[ n(A \times B) = n(A) \times n(B) \]
where \( n(A) \) is the number of elements in set A and \( n(B) \) is the number of elements in set B.

### Step 2: Given Information
We know that:
- \( n(A \times B) = 6 \)
- The elements of \( A \times B \) are: \( (1, 2), (2, 3), (3, 3) \)

### Step 3: Determine the Elements of Sets A and B
From the given elements of \( A \times B \):
- The first elements (from set A) are: 1, 2, 3
- The second elements (from set B) are: 2, 3, 3

From this, we can deduce:
- Set A = {1, 2, 3}
- Set B = {2, 3}

### Step 4: Verify the Number of Elements
Now we can find the number of elements in each set:
- \( n(A) = 3 \) (since A has three elements: 1, 2, 3)
- \( n(B) = 2 \) (since B has two elements: 2, 3)

Now, check the product:
\[ n(A \times B) = n(A) \times n(B) = 3 \times 2 = 6 \]
This matches the given information.

### Step 5: Find the Cartesian Product \( A \times B \)
Now, we will list all the ordered pairs for \( A \times B \):
- For 1 from A: (1, 2), (1, 3)
- For 2 from A: (2, 2), (2, 3)
- For 3 from A: (3, 2), (3, 3)

Thus,
\[ A \times B = \{(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)\} \]

### Step 6: Find the Cartesian Product \( B \times A \)
Now, we will list all the ordered pairs for \( B \times A \):
- For 2 from B: (2, 1), (2, 2), (2, 3)
- For 3 from B: (3, 1), (3, 2), (3, 3)

Thus,
\[ B \times A = \{(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\} \]

### Final Answer
- \( A \times B = \{(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)\} \)
- \( B \times A = \{(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\} \)