If `n (AnnB)=10, n (BnnC) =20 and n(AnnC)=30,` then the greatest possible value of `n(AnnBnnC)` is

Options

A. 15
B. 20
C. 10
D. 4

Answer

To find the greatest possible value of \( n(A \cap B \cap C) \), we can use the information provided about the intersections of sets A, B, and C.

Given:
- \( n(A \cap B) = 10 \)
- \( n(B \cap C) = 20 \)
- \( n(A \cap C) = 30 \)

We want to determine the maximum value of \( n(A \cap B \cap C) \).

### Step-by-Step Solution:

1. **Understanding the Intersections**:
- The intersection \( A \cap B \) contains elements that are in both sets A and B. Since \( n(A \cap B) = 10 \), this means there are 10 elements common to both A and B.
- The intersection \( B \cap C \) contains elements that are in both sets B and C. Since \( n(B \cap C) = 20 \), there are 20 elements common to both B and C.
- The intersection \( A \cap C \) contains elements that are in both sets A and C. Since \( n(A \cap C) = 30 \), there are 30 elements common to both A and C.

2. **Finding the Maximum Intersection**:
- The intersection \( A \cap B \cap C \) is the set of elements that are common to all three sets A, B, and C.
- To find the greatest possible value of \( n(A \cap B \cap C) \), we need to consider the constraints provided by the sizes of the intersections.

3. **Using the Minimum Value**:
- The maximum number of elements that can be in \( A \cap B \cap C \) cannot exceed the smallest intersection among \( A \cap B \), \( B \cap C \), and \( A \cap C \).
- Therefore, we take the minimum of the values:
\[ \text{Minimum}(n(A \cap B), n(B \cap C), n(A \cap C)) = \text{Minimum}(10, 20, 30) = 10 \]

4. **Conclusion**:
- The greatest possible value of \( n(A \cap B \cap C) \) is 10.

Thus, the final answer is:
\[ \boxed{10} \]