Question

Let awards in event A is 48 and awards in event B is 25 and awards in event C is 18 and also n(A ∪ B ∪ C) = 60, n(A ⋂ B ⋂ C) = 5, then how many got exactly two awards is?

• A. 21
• B. 25
• C. 24
• D. 23

Answer 1

The Correct Option is A

To solve the problem of determining how many students received exactly two awards among events A, B, and C, we can use the principle of inclusion-exclusion. Let's define:

• \(n(A) = 48\)
• \(n(B) = 25\)
• \(n(C) = 18\)
• \(n(A \cup B \cup C) = 60\)
• \(n(A \cap B \cap C) = 5\)

The goal is to find the number of students who received exactly two awards. Let's denote:

• \(x = n((A \cap B) \setminus C)\)
• \(y = n((A \cap C) \setminus B)\)
• \(z = n((B \cap C) \setminus A)\)

The formula for inclusion-exclusion is:

\(n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)\)

Given the information:

\(60 = 48 + 25 + 18 - n(A \cap B) - n(A \cap C) - n(B \cap C) + 5\)

Simplifying, we get:

\(60 = 91 - (n(A \cap B) + n(A \cap C) + n(B \cap C)) + 5\)

\(60 = 96 - (n(A \cap B) + n(A \cap C) + n(B \cap C))\)

\(n(A \cap B) + n(A \cap C) + n(B \cap C) = 36\)

Now, calculate the exact number of students who got exactly two awards:

The number of students who got exactly two awards without being part of all three awards is:

\(x + y + z = (n(A \cap B) - 5) + (n(A \cap C) - 5) + (n(B \cap C) - 5)\)

\(= n(A \cap B) + n(A \cap C) + n(B \cap C) - 3 \times 5\)

\(= 36 - 15\)

\(= 21\)

Thus, the number of students who got exactly two awards is \(21\).