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

Let the total number of men be represented by the set \( A \cup B \cup C = 60 \) where:
- \( |A| = 48 \) (men who received medals in event A)
- \( |B| = 25 \) (men who received medals in event B)
- \( |C| = 18 \) (men who received medals in event C)
- \( |A \cup B \cup C| = 60 \) (total number of men)
The number of men who received medals in all three events is given by: \[ |A \cap B \cap C| = 5 \]
We need to find how many men received medals in exactly two events, which is calculated by: \[ |A \cap B| + |B \cap C| + |C \cap A| - 3|A \cap B \cap C| \] Using the inclusion-exclusion principle, we get: \[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C| \] Substituting the values we know: \[ 60 = 48 + 25 + 18 - |A \cap B| - |B \cap C| - |C \cap A| + 5 \] \[ 60 = 91 - |A \cap B| - |B \cap C| - |C \cap A| + 5 \] \[ |A \cap B| + |B \cap C| + |C \cap A| = 36 \] Now, to find the number of men who received exactly two medals, we use the formula: \[ \text{No. of men who received exactly 2 medals} = |A \cap B| + |B \cap C| + |C \cap A| - 3|A \cap B \cap C| \] Substituting the values: \[ \text{No. of men who received exactly 2 medals} = 36 - 15 = 21 \] Thus, the number of men who received exactly two medals is 21.