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

Answer 1

Answer: (D)

Any of the three values 26, 27, 28 is possible

Answer 2

We have $n \left(A \cup B \cup C\right) = n \left(A\right) + n \left(B\right) + n \left(C\right) -$ $n \left(A \cap B\right) - n\left(B\cap C\right) - n \left(C \cap A\right) + n \left(A\cap B \cap C\right)$ $= 10 +15 + 20 - 8 - 9 - n \left(C \cap A\right) + n \left(A \cap B \cap C\right)$ $= 28 - n\left(C \cap A\right) - n \left(A \cap B \cap C\right)\quad ...\left(i\right)$ Since $n \left(C \cap A\right) \ge n \left(A \cap B \cap C\right)$ We have $n \left(C \cap A\right) - n \left(A \cap B \cap C\right) \ge 0\quad...\left(ii\right)$ From $\left(i\right)$ and $\left(ii\right)$ $n \left(A \cup B \cup C\right) \le 28\quad ...\left(iii\right)$ Now, $n\left(A \cup B\right) = n \left(A\right) +n \left(B\right) - n \left(A \cap B\right)$ $= 10 + 15 - 8 = 17$ and $n \left(B \cup C\right) = n \left(B\right) + n \left(C\right) - n \left(B \cap C\right)$ $= 15 + 20 - 9 = 26$ Since, $n \left(A \cup B \cup C\right) \ge n \left(A\cup C\right)$ and $n \left(A\cup B\cup C\right) \ge n \left(B\cup C\right)$, we have $n \left(A\cup B\cup C\right) \ge 17$ and $n \left(A\cup B\cup C\right) \ge 26$ Hence $n \left(A\cup B\cup C\right) \ge 26 \quad...\left(iv\right)$ From $\left(iii\right)$ and $\left(iv\right)$ we obtain $26 \le n \left(A\cup B\cup C\right) \le 28$ Also $n \left(A\cup B\cup C\right)$ is a positive integer $\therefore n\left(A\cup B\cup C\right) = 26$ or $27$ or $28$