If `n (A)=8 and n (A nnB)=2,` then `n[(AnnB)' nn A]` is equal to

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem, we need to find the value of \( n[(A \cap B)' \cap A] \) given that \( n(A) = 8 \) and \( n(A \cap B) = 2 \).

### Step-by-Step Solution:

1. **Understand the Notation**:
- \( n(A) \) represents the number of elements in set \( A \).
- \( n(A \cap B) \) represents the number of elements in the intersection of sets \( A \) and \( B \).
- \( (A \cap B)' \) represents the complement of the intersection of sets \( A \) and \( B \).

2. **Calculate \( n(A \cap B)' \)**:
- The complement of \( A \cap B \) includes all elements in \( A \) that are not in \( B \).
- The number of elements in \( A \cap B \) is given as 2.
- Therefore, the number of elements in \( (A \cap B)' \) can be calculated as:
\[ n((A \cap B)') = n(A) - n(A \cap B) = 8 - 2 = 6 \]

3. **Find \( n[(A \cap B)' \cap A] \)**:
- The intersection \( (A \cap B)' \cap A \) includes elements that are in \( A \) but not in \( B \).
- Since we already calculated that \( n((A \cap B)') = 6 \), and since all these elements are in \( A \), we find:
\[ n[(A \cap B)' \cap A] = n((A \cap B)') = 6 \]

Thus, the final answer is:
\[ n[(A \cap B)' \cap A] = 6 \]