JEE Mathematics 2023 Question with Solution

Given:

• $|A| = 48$
• $|B| = 25$
• $|C| = 18$
• $|A \cup B \cup C| = 60$
• $|A \cap B \cap C| = 5$

Find: Number of men who received medals in exactly two events.

Using the inclusion-exclusion principle,

Substituting the given values,

So,

The number of men who received medals in exactly two events is

Therefore,

Therefore, the number of men who received medals in exactly two of the three events is $21$. The correct option is C.

Given: The sum $|A \cap B| + |B \cap C| + |C \cap A|$ counts every man who is in all three events exactly $3$ times, once in each pairwise intersection.

Find: Why the formula for exactly two events subtracts $3|A \cap B \cap C|$.

A man who got medals in exactly two events appears in exactly one of the pairwise intersections. But a man who got medals in all three events appears in

so he is counted $3$ times.

Hence, to keep only those who are in exactly two events, we subtract all three of those counts:

Using

we get

So the required number is $21$.