Let A, B , C be finite sets. Suppose that n `(A) = 10, n (B) = 15, n (C) = 20 , n (A nn B) = 8 and n (B nn C)=9.` Then the possible value of a `(A uu B uu C)` is

Options

A. 26
B. 27
C. 28
D. Any of the three values 26, 27, 28 is possible

Answer

To solve the problem, we need to find the possible values of \( n(A \cup B \cup C) \) given the sizes of the sets and their intersections. We will use the principle of inclusion-exclusion.

### Step-by-Step Solution:

1. **Identify Given Values:**
- \( n(A) = 10 \)
- \( n(B) = 15 \)
- \( n(C) = 20 \)
- \( n(A \cap B) = 8 \)
- \( n(B \cap C) = 9 \)

2. **Use the Inclusion-Exclusion Principle:**
The formula for the union of three sets is:
\[ n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) \]
We need to express \( n(A \cup B \cup C) \) in terms of the known values and unknowns \( n(A \cap C) \) and \( n(A \cap B \cap C) \).

3. **Substituting Known Values:**
Substitute the known values into the formula:
\[ n(A \cup B \cup C) = 10 + 15 + 20 - 8 - 9 - n(A \cap C) + n(A \cap B \cap C) \]
Simplifying this gives:
\[ n(A \cup B \cup C) = 28 - n(A \cap C) + n(A \cap B \cap C) \]

4. **Establish Relationships:**
- From the intersection sizes, we know that:
\[ n(A \cap B) \geq n(A \cap B \cap C) \]
This implies \( 8 \geq n(A \cap B \cap C) \).
- Also, since \( n(A \cap C) \) cannot exceed \( n(A) \) or \( n(C) \), we have:
\[ n(A \cap C) \leq 10 \quad \text{and} \quad n(A \cap C) \leq 20 \]

5. **Finding Bounds for \( n(A \cup B \cup C) \):**
- To find the minimum value of \( n(A \cup B \cup C) \), we can assume \( n(A \cap C) \) is at its maximum (10) and \( n(A \cap B \cap C) \) is at its minimum (0):
\[ n(A \cup B \cup C) \geq 28 - 10 + 0 = 18 \]
- To find the maximum value, we can assume \( n(A \cap C) \) is at its minimum (0) and \( n(A \cap B \cap C) \) is at its maximum (8):
\[ n(A \cup B \cup C) \leq 28 - 0 + 8 = 36 \]

6. **Finding Possible Values:**
- Now we need to determine the possible integer values for \( n(A \cup B \cup C) \) between the bounds we calculated.
- Since \( n(A \cap C) \) can vary from 0 to 10, and \( n(A \cap B \cap C) \) can vary from 0 to 8, we can find that:
- The minimum possible value of \( n(A \cup B \cup C) \) is 26 (when \( n(A \cap C) = 10 \) and \( n(A \cap B \cap C) = 0 \)).
- The maximum possible value is 28 (when \( n(A \cap C) = 0 \) and \( n(A \cap B \cap C) = 8 \)).
- Therefore, the possible values of \( n(A \cup B \cup C) \) are 26, 27, and 28.

### Conclusion:
The possible values of \( n(A \cup B \cup C) \) are 26, 27, and 28.