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 solve the problem, we need to find the greatest possible value of \( n(A \cap B \cap C) \) given the following information:

1. \( n(A \cap B) = 10 \)
2. \( n(B \cap C) = 20 \)
3. \( n(A \cap C) = 30 \)

### Step-by-Step Solution:

**Step 1: Understanding the Problem**
We are given the sizes of the intersections of three sets A, B, and C. We need to find the maximum number of elements that can be in the intersection of all three sets, \( n(A \cap B \cap C) \).

**Hint for Step 1:** Think about how the intersections relate to each other and what it means for elements to be in multiple sets.

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**Step 2: Applying the Principle of Inclusion-Exclusion**
We know that:
- \( n(A \cap B) \) represents the total number of elements that are in both A and B.
- \( n(B \cap C) \) represents the total number of elements that are in both B and C.
- \( n(A \cap C) \) represents the total number of elements that are in both A and C.

To find \( n(A \cap B \cap C) \), we need to consider the overlaps among these intersections.

**Hint for Step 2:** Consider how many elements can be counted in each intersection without exceeding the total counts given.

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**Step 3: Finding the Maximum Value**
Since \( n(A \cap B) = 10 \), this means that at most 10 elements can be in both A and B. Therefore, the maximum number of elements that can also be in C (i.e., \( n(A \cap B \cap C) \)) cannot exceed 10.

**Hint for Step 3:** The maximum value of \( n(A \cap B \cap C) \) is limited by the smallest intersection size among \( n(A \cap B) \), \( n(B \cap C) \), and \( n(A \cap C) \).

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**Step 4: Conclusion**
Thus, the greatest possible value of \( n(A \cap B \cap C) \) is:
\[ \text{Greatest possible value of } n(A \cap B \cap C) = 10 \]

### Final Answer:
The greatest possible value of \( n(A \cap B \cap C) \) is **10**.

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