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 this problem, we will use the principle of inclusion-exclusion which is often used in set theory to find the number of elements in the union of multiple sets.

Given the following data:

• Awards in event A: \( n(A) = 48 \)
• Awards in event B: \( n(B) = 25 \)
• Awards in event C: \( n(C) = 18 \)
• Total awards in union of events A, B, and C: \( n(A \cup B \cup C) = 60 \)
• Awards in all three events: \( n(A \cap B \cap C) = 5 \)

We need to find the number of participants who got exactly two awards. Let's denote:

• \( x = n(A \cap B) \) (number of awards in both A and B)
• \( y = n(B \cap C) \) (number of awards in both B and C)
• \( z = n(C \cap A) \) (number of awards in both C and A)

Using the principle of inclusion-exclusion:

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

Substituting the known values:

\(60 = 48 + 25 + 18 - x - y - z + 5\)

\(60 = 91 - x - y - z + 5\)

Simplifying further:

\(60 = 96 - x - y - z\)

\(x + y + z = 96 - 60 = 36\)

Now, to find the number of participants who got exactly two awards, subtract those who got all three:

\(x + y + z - 3 \times n(A \cap B \cap C) = 36 - 3 \times 5 = 36 - 15 = 21\)

Therefore, the number of participants who received exactly two awards is 21.