In a class of 140 students numbered 1 to 140, all even numbered students opted mathematics course, t

Q: Wait, why do we use the floor function $\lfloor \frac{140}{n} \rfloor$ instead of just dividing?

Great question! In set theory problems involving divisibility, we are counting discrete items—students. If you divide $140$ by $3$, you get $46.66...$. Since you cannot have $0.66$ of a student, the floor function $\lfloor \cdot \rfloor$ effectively rounds down to the nearest whole number, giving us the exact count of students whose roll numbers are multiples of $3$.

Q: Why do we add the triple intersection $n(M \cap P \cap C)$ back at the end of the formula?

Think of it as a correction. When we added the individual sets, we counted the triple-intersection students three times. When we subtracted the double-intersections, we removed them three times. So, they are currently not counted at all! We add them back once to ensure they are included in the final union count.

Q: Can I just subtract the intersections from the total number of students directly?

No, that is a common trap! You must first find the union $n(M \cup P \cup C)$ using the Inclusion-Exclusion Principle. Only after you have the total number of students who took at least one course can you subtract that value from the universal set of $140$ to find those who took none.

Question

In a class of 140 students numbered 1 to 140, all even numbered students opted mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is :

Options:

(A) 102

(B) 42

(C) 1

(D) 38

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a universal set of $n(U) = 140$ students, each assigned a unique roll number from $1$ to $140$. We define three sets based on divisibility:
*   Math ($M$): Numbers divisible by $2$.
*   Physics ($P$): Numbers divisible by $3$.
*   Chemistry ($C$): Numbers divisible by $5$.

Our goal is to find the number of students who opted for none of these subjects, which is represented by $n(U) - n(M \cup P \cup C)$.

### Phase 1: The Individual Circles

First, we calculate the cardinality of each individual set using the floor function:

$$n(M) = \lfloor \frac{140}{2} floor = 70$$

$$n(P) = \lfloor \frac{140}{3} floor = 46$$

$$n(C) = \lfloor \frac{140}{5} floor = 28$$

These values represent the total count of students in each subject, but they currently include students who are enrolled in multiple subjects.

### Phase 2: The Overlapping Realities

To correct for multiple counting, we identify the intersections of these sets by finding the Least Common Multiple (LCM) of the divisors:

$$n(M \cap P) = \lfloor \frac{140}{	ext{lcm}(2,3)} floor = \lfloor \frac{140}{6} floor = 23$$

$$n(P \cap C) = \lfloor \frac{140}{	ext{lcm}(3,5)} floor = \lfloor \frac{140}{15} floor = 9$$

$$n(M \cap C) = \lfloor \frac{140}{	ext{lcm}(2,5)} floor = \lfloor \frac{140}{10} floor = 14$$

### Phase 3: The Triple Intersection and the Grand Formula

We must also account for the students enrolled in all three subjects, whose roll numbers are divisible by $	ext{lcm}(2, 3, 5) = 30$:

$$n(M \cap P \cap C) = \lfloor \frac{140}{30} floor = 4$$

Now, we apply the **Principle of Inclusion-Exclusion** to find the union of the three sets:

$$n(M \cup P \cup C) = n(M) + n(P) + n(C) - [n(M \cap P) + n(P \cap C) + n(M \cap C)] + n(M \cap P \cap C)$$

Substituting the calculated values:

$$n(M \cup P \cup C) = 70 + 46 + 28 - (23 + 9 + 14) + 4$$

$$n(M \cup P \cup C) = 144 - 46 + 4 = 102$$

### Final Calculation

The value $102$ represents the number of students who opted for at least one subject. To find the number of students who opted for none, we subtract this from the universal set:

$$	ext{Result} = n(U) - n(M \cup P \cup C) = 140 - 102 = 38$$

The number of students who opted for none of the subjects is **38**.

Select Answer:

Visualized Solution

Visualizing the Problem

Counting Mathematics Students ( $M$ )

Counting Physics Students ( $P$ )

Counting Chemistry Students ( $C$ )

Finding Intersections: $M \cap P$

Finding Intersections: $P \cap C$

Finding Intersections: $M \cap C$

The Triple Intersection: $M \cap P \cap C$

Principle of Inclusion-Exclusion

Calculating the Union

Final Answer: Students with No Course

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

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