In a survey of 220 students of a higher secondary school, it was found that at least 125 and at most

Q: Why do we subtract 10 from 220? Isn't the total number of students 220?

Great catch! The 220 represents the entire population of the survey. However, the Principle of Inclusion-Exclusion operates on the union of the sets (those who study at least one subject). Since 10 students study nothing, they fall outside our Venn diagram circles. We must focus on the 210 students who actually occupy the space within the sets.

Q: Why can't $x$ be larger than 30? I calculated the range as 15 to 45.

Mathematical bounds are necessary but not sufficient! While the algebra allows $x$ up to 45, we must respect the physical reality of the sets. The triple intersection $n(M \cap P \cap C)$ is a subset of every pairwise intersection. Since only 30 students study both Physics and Chemistry, it is impossible for more than 30 to study all three. We must intersect our algebraic range with this physical constraint.

Q: How do I know if I should add or subtract the ranges for $n(M) + n(P) + n(C)$?

Think of it as a balance scale. To find the minimum possible value of the sum, you take the minimum of each individual component. To find the maximum, you take the maximum of each. Since the sum is bounded by the individual constraints, adding the lower bounds gives the lower limit of the sum, and adding the upper bounds gives the upper limit.

Q: Is it possible for $x$ to be 0?

In this specific problem, no. Our algebraic derivation $15 \le x \le 45$ forces $x$ to be at least 15. If your calculation ever leads to a range that includes zero, you would then check if the problem constraints allow for an empty intersection. Always trust the algebra, but verify it against the logic of the sets.

Q: Why is the Principle of Inclusion-Exclusion so important here?

It is the bridge between the 'parts' and the 'whole'. Without it, we would be double-counting (or triple-counting) students who study multiple subjects. It allows us to treat the complex overlapping regions as a single, coherent equation, turning a word problem into a solvable algebraic inequality.

Question

In a survey of 220 students of a higher secondary school, it was found that at least 125 and at most 130 students studied Mathematics; at least 85 and at most 95 studied Physics; at least 75 and at most 90 studied Chemistry; 30 studied both Physics and Chemistry; 50 studied both Chemistry and Mathematics; 40 studied both Mathematics and Physics and 10 studied none of these subjects. Let $m$ and $n$ respectively be the least and the most number of students who studied all the three subjects. Then $m + n$ is equal to ______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the middle of a bustling school hallway. You have a clipboard, and you are tasked with understanding the academic lives of $220$ students.

Some love Mathematics, some are drawn to the elegance of Physics, and others find their passion in the chemical reactions of Chemistry. But the world is rarely simple—students overlap, interests collide, and some students might not be studying any of these at all.

This is the essence of the **Principle of Inclusion-Exclusion**, and today, we are going to master it.

### Defining the Universe

First, let us define our universe. We have a total of $N = 220$ students.

However, the survey tells us that $10$ students study none of these subjects. This is our first crucial step: we must isolate the students who actually participate in our sets.

The union of our three sets, $n(M \cup P \cup C)$, is simply the total population minus those who opted out: $210$. Think of this as clearing the stage before the performance begins.

### The Master Equation

We are looking for $x$, the number of students who study all three subjects—the heart of the Venn diagram. We know the pairwise intersections: $n(M \cap P) = 40$, $n(P \cap C) = 30$, and $n(C \cap M) = 50$.

The **Principle of Inclusion-Exclusion** is our most powerful tool here. It states:

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

By substituting our known values, we get:

$$210 = n(M) + n(P) + n(C) - (40 + 30 + 50) + x$$

Simplifying this, we arrive at the elegant relationship:

$$n(M) + n(P) + n(C) + x = 330$$

This equation is the heartbeat of our problem. It tells us that the sum of the individual sets and the triple intersection is a constant $330$. If one group grows, the others must shift to compensate.

### The Dance of Inequalities

Now, we introduce the constraints. We know that $125 \le n(M) \le 130$, $85 \le n(P) \le 95$, and $75 \le n(C) \le 90$.

To find the range of the sum $n(M) + n(P) + n(C)$, we simply add the lower bounds and the upper bounds:

$$(125 + 85 + 75) \le n(M) + n(P) + n(C) \le (130 + 95 + 90)$$

This gives us $285 \le n(M) + n(P) + n(C) \le 315$. Since we know $n(M) + n(P) + n(C) = 330 - x$, we can substitute this into our inequality:

$$285 \le 330 - x \le 315$$

Solving this for $x$ is like tightening a vice. Subtracting $330$ from all sides, we get $-45 \le -x \le -15$. Multiplying by $-1$ (and flipping the inequality signs), we find $15 \le x \le 45$.

### The Final Reality Check

We are almost there, but we must be careful. Mathematics is not just about symbols; it is about logic.

The triple intersection $x$ cannot be larger than any of the pairwise intersections. If only $30$ students study both Physics and Chemistry, it is physically impossible for $45$ students to study all three.

Thus, we must enforce the constraint $x \le 30$.

Combining our algebraic range $15 \le x \le 45$ with our physical constraint $x \le 30$, we find the true range: $15 \le x \le 30$.

Our least value $m$ is $15$, and our maximum value $n$ is $30$. The final answer, **$m + n = 45$**, is not just a number—it is the culmination of understanding how these overlapping sets define the structure of the student body.

Enter Numerical Value:

Visualized Solution

Visualizing the Survey Data

Finding the Union $n (M \cup P \cup C)$

Defining the Central Intersection $x$

Setting up Pairwise Intersections

Applying Inclusion-Exclusion Principle

Substituting Known Values

Simplifying the Master Equation

Analyzing the Given Ranges

Finding the Combined Range

Bounding the Variable $x$

Solving for $x$ Limits

The Hidden Intersection Constraint

Final Range of $x$

Calculating $m + n$

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

Defining the Universe

The Master Equation

The Dance of Inequalities

The Final Reality Check

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