The set A and B has 3 common elements . If n(A) =5 and n(B)=4, then find `n(AxxB)` and `x{(AxxB)nn (BxxA)}`

Answer

To solve the problem, we need to find \( n(A \times B) \) and \( n((A \times B) \cap (B \times A)) \).

### Step 1: Understand the given information
We know:
- \( n(A) = 5 \) (the number of elements in set A)
- \( n(B) = 4 \) (the number of elements in set B)
- The number of common elements in sets A and B, denoted as \( n(A \cap B) = 3 \).

### Step 2: Calculate \( n(A \times B) \)
The Cartesian product \( A \times B \) consists of all ordered pairs \( (a, b) \) where \( a \in A \) and \( b \in B \). The number of elements in the Cartesian product is given by the formula:
\[ n(A \times B) = n(A) \times n(B) \]
Substituting the values:
\[ n(A \times B) = 5 \times 4 = 20 \]

### Step 3: Calculate \( n(B \times A) \)
Similarly, the number of elements in \( B \times A \) is:
\[ n(B \times A) = n(B) \times n(A) = 4 \times 5 = 20 \]

### Step 4: Calculate \( n((A \times B) \cap (B \times A)) \)
To find the intersection \( (A \times B) \cap (B \times A) \), we need to consider the pairs that are common in both products. The common pairs will be of the form \( (x, x) \) where \( x \) is an element in \( A \cap B \).

Since \( n(A \cap B) = 3 \), there are 3 common elements in both sets. Therefore, the number of common pairs will be:
\[ n((A \times B) \cap (B \times A)) = n(A \cap B) = 3 \]

### Final Results
- \( n(A \times B) = 20 \)
- \( n((A \times B) \cap (B \times A)) = 3 \)