Question

An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to total 60 men and only five men got medals in all the three events, then how many received medals in exactly two of three events?

• A. 15
• B. 9
• C. 21
• D. 10

Hint:

To solve inclusion-exclusion problems, always remember to account for the overlap of sets, subtracting the number of elements in the intersection of all sets when required.

Answer 1

The Correct Option is C

To solve this problem, we can use the principle of inclusion-exclusion to determine how many men received medals in exactly two of the three events.

First, let's define the sets:

• \(A\): The set of men who received medals in event 'A'.
• \(B\): The set of men who received medals in event 'B'.
• \(C\): The set of men who received medals in event 'C'.

Given:

• \(|A| = 48\)
• \(|B| = 25\)
• \(|C| = 18\)
• Total men who received medals in any event, \(|A \cup B \cup C| = 60\)
• The number of men who received medals in all three events, \(|A \cap B \cap C| = 5\)

We need to find how many men received medals in exactly two events, denoted as \(|A \cap B| + |B \cap C| + |C \cap A| - 3|A \cap B \cap C|\).

Using the principle of inclusion-exclusion:

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

Substitute the known values:

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

Solving for \(|A \cap B| + |B \cap C| + |C \cap A|\):

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

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

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

Now, calculate the number of men who received medals in exactly two events:

\(|A \cap B| + |B \cap C| + |C \cap A| - 3|A \cap B \cap C| = 36 - 3 \times 5 = 36 - 15 = 21\)

Hence, the number of men who received medals in exactly two of the three events is 21.