If `n(AcupB)=18 `n(A cap B)=5`,n(B -A)=3`,then find `n(A cap B)`

Answer

To solve the problem step by step, we will use the information given in the question and apply the principles of set theory.

### Step 1: Understand the given information
We are given:
- \( n(A \cup B) = 18 \) (the number of elements in the union of sets A and B)
- \( n(A - B) = 5 \) (the number of elements in A that are not in B)
- \( n(B - A) = 3 \) (the number of elements in B that are not in A)

We need to find \( n(A \cap B) \) (the number of elements common to both sets A and B).

### Step 2: Use the formula for the union of two sets
The formula for the number of elements in the union of two sets is given by:
\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]

### Step 3: Express \( n(A) \) and \( n(B) \) in terms of \( n(A \cap B) \)
From the definitions of set differences:
- \( n(A) = n(A - B) + n(A \cap B) \)
- \( n(B) = n(B - A) + n(A \cap B) \)

Substituting the known values:
- \( n(A) = 5 + n(A \cap B) \)
- \( n(B) = 3 + n(A \cap B) \)

### Step 4: Substitute into the union formula
Now we substitute \( n(A) \) and \( n(B) \) into the union formula:
\[ n(A \cup B) = (5 + n(A \cap B)) + (3 + n(A \cap B)) - n(A \cap B) \]
This simplifies to:
\[ n(A \cup B) = 5 + 3 + n(A \cap B) = 8 + n(A \cap B) \]

### Step 5: Set up the equation
Now we know that:
\[ n(A \cup B) = 18 \]
So we set up the equation:
\[ 8 + n(A \cap B) = 18 \]

### Step 6: Solve for \( n(A \cap B) \)
To find \( n(A \cap B) \), we subtract 8 from both sides:
\[ n(A \cap B) = 18 - 8 \]
\[ n(A \cap B) = 10 \]

### Final Answer
Thus, the number of elements in \( A \cap B \) is:
\[ \boxed{10} \]