Question

Let \(A, B, C\) be finite sets. Suppose that \(n(A) = 10\), \(n(B) = 15\), \(n(C) = 20\), \(n(A \cap B) = 8\), and \(n(B \cap C) = 9\). Then the possible value of \(n(A \cup B \cup C)\) is

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

Hint:

If intersections are not fully specified, multiple answers may be possible.

Answer 1

The Correct Option is D

Step 1:
\[ n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) \]

Step 2: Since \(n(A \cap C)\) and \(n(A \cap B \cap C)\) are not fixed, multiple values are possible.

Step 3: Valid arrangements give 26, 27, or 28.