A class has 175 students. The following data shows the number of students obtaining one or more subjects. Mathematics 100 , Physics 70 , Chemistry 40; Mathematics and Physics 30 , Mathematics and Chemistry 28 , Physics and Chemistry 23; Mathematics, Physics and Chemistry 18 How many students have offered Mathematics alone?

1. What Concept Is Tested?

The problem is a direct application of the Principle of Inclusion–Exclusion (PIE) for three finite sets.

For three sets $M,\,P,\,C$ we have

$|M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |M \cap C| - |P \cap C| + |M \cap P \cap C|.$

If we want exactly one set (say only $M$), we subtract the two‐way and add the three‐way intersections, because the three‐way part had been subtracted twice.

[ \text{Only }M = |M| - |M \cap P| - |M \cap C| + |M \cap P \cap C| ]

2. Logical Flow a Student Should Follow

1. Identify the Sets: $M$ (Math), $P$ (Physics), $C$ (Chemistry).
2. Write Down Given Cardinalities.
3. Recognise the Need for PIE because counts of overlaps are supplied.
4. Plug into the formula for ‘only‐one’ sets.
5. Compute, keeping signs correct (− for two‐way, + for three‐way).
6. Box the Answer.

3. Why Each Step Makes Sense

• PIE prevents double- or triple-counting.
• Subtracting two-way overlaps removes students who have an extra subject.
• Adding back the three-way overlap corrects for over-subtraction.

That’s the entire reasoning chain for this straightforward JEE‐level set problem.

Imagine Three Big Circles in a Playground

Think of Mathematics (M), Physics (P) and Chemistry (C) as three big circles painted on the ground. Every student stands in the circle(s) for the subject(s) they have chosen.

We know how many students are in:

• The whole Math circle (100)
• The whole Physics circle (70)
• The whole Chemistry circle (40)
• The overlaps of two circles (Math & Physics = 30, Math & Chemistry = 28, Physics & Chemistry = 23)
• The tiny spot where all three circles overlap (Math & Physics & Chemistry = 18)

To find “Math‐only” kids, we start with all 100 standing somewhere in the Math circle and simply kick out everyone who is also touching Physics or Chemistry. But notice: the kids touching both extra circles (all three subjects) will be kicked out twice, so we have to invite them back once!

So Math‐only kids = 100 − 30 − 28 + 18 = 60 students.

Step-by-Step Solution

Let

• $|M| = 100$
• $|P| = 70$
• $|C| = 40$
• $|M \cap P| = 30$
• $|M \cap C| = 28$
• $|P \cap C| = 23$
• $|M \cap P \cap C| = 18$

We want the number of students who chose Mathematics only: