A boy needs to select five courses from 12 available courses, out of which 5 courses are language co

Q: Wait, why can't I just pick 2 language courses first and then pick 3 more from the remaining 10 courses? That seems faster.

That is a classic trap! If you pick 2 language courses first and then 3 from the remaining 10, you are implicitly assuming that the order of selection matters or that you are overcounting specific combinations. By fixing the number of language courses in each case, we ensure that every combination is counted exactly once, without overlap. Always partition your sample space when constraints are involved!

Q: Why do we add the results of the cases? Shouldn't we multiply them?

Great question. We multiply *within* a case because we are performing a sequence of independent choices (choosing language AND non-language courses). We add *between* cases because these scenarios are mutually exclusive—you cannot simultaneously have 'zero language courses' and 'one language course'. The Addition Principle is for 'OR' logic, and the Multiplication Principle is for 'AND' logic.

Q: Is there a faster way to solve this using subtraction, like total ways minus the unwanted cases?

Yes, you could calculate the total ways to choose 5 courses from 12 (${}^{12}C_5$) and subtract the cases where you have 3, 4, or 5 language courses. However, in this specific problem, calculating the 3, 4, and 5 language course scenarios takes roughly the same amount of effort as calculating the 0, 1, and 2 scenarios. Always choose the path that minimizes your calculation steps to avoid arithmetic errors!

Question

A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is ____.

Sigma Percentile · Accepted Answer

### The Art of Selection: Mastering Combinatorial Constraints

Welcome, future engineer. Today, we aren't just solving a combinatorics problem; we are learning how to organize chaos.

When you look at a problem like this, it is easy to feel overwhelmed by the sheer number of possibilities. But the secret to JEE Advanced success isn't brute force—it is the ability to partition a complex problem into manageable, elegant pieces.

### Phase 1: Deconstructing the Constraint

We are presented with a pool of $12$ courses: $5$ language courses and $7$ non-language courses. We need to select $5$ courses total.

The constraint is the gatekeeper here: 'at most two' language courses.

Imagine you are standing in front of a course registration desk. The clerk tells you, 'You can take zero, one, or two language courses, but no more.'

This is your signal to break the problem into three mutually exclusive worlds. We don't try to solve for everything at once; we solve for Case $0$, Case $1$, and Case $2$, and then we bring them together.

### Phase 2: The Three Worlds

Let's step into these worlds one by one.

**Case 1: The Non-Linguist (Zero Language Courses)**

If we choose $0$ language courses, all $5$ of our selections must come from the $7$ non-language courses. Mathematically, this is represented as:

$$^{5}C_{0} 	imes ^{7}C_{5}$$

Since $^{5}C_{0} = 1$ and $^{7}C_{5}$ is equivalent to $^{7}C_{2}$, we calculate:

$$\frac{7 	imes 6}{2 	imes 1} = 21$$

So, there are $21$ ways to walk this path.

**Case 2: The Balanced Scholar (One Language Course)**

What if we decide to take exactly one language course? We pick $1$ from the $5$ available language courses, and the remaining $4$ must come from the $7$ non-language courses.

This gives us:

$$^{5}C_{1} 	imes ^{7}C_{4}$$

With $^{5}C_{1} = 5$ and $^{7}C_{4}$ (which is the same as $^{7}C_{3}$) equaling $35$, we find:

$$5 	imes 35 = 175 	ext{ ways}$$

**Case 3: The Maximum Limit (Two Language Courses)**

Finally, we reach the limit. We select $2$ language courses from the $5$ available, and $3$ non-language courses from the $7$ available.

This is expressed as:

$$^{5}C_{2} 	imes ^{7}C_{3}$$

We know $^{5}C_{2} = 10$ and $^{7}C_{3} = 35$. Multiplying these gives us:

$$10 	imes 35 = 350 	ext{ ways}$$

### Phase 3: The Grand Unification

Now, we have our three distinct counts: $21$, $175$, and $350$. Because these cases are mutually exclusive—meaning you cannot be in Case $1$ and Case $2$ at the same time—we use the Addition Principle.

$$Total = 21 + 175 + 350 = 546$$

There it is. **$546$ ways**.

It is not just a number; it is the result of systematic, logical thinking. When you face these problems in the exam hall, remember this: don't panic at the constraint. Break it down, calculate the pieces, and trust the process.

Animated Solution for Mathematics - Permutations and Combinations: A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is ____.

Enter Numerical Value:

Visualized Solution

Understanding the Course Pool

Analyzing the Constraint

Case 1 Setup (Zero Language Courses)

Case 1 Calculation

Case 2 Setup (One Language Course)

Case 2 Calculation

Case 3 Setup (Two Language Courses)

Case 3 Calculation

Total Number of Ways

The Sigma Insight: Combinations and Selection

The Art of Selection

Phase 1

Phase 2

Phase 3

Similar Questions

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

There are ten boys $B_{1}, B_{2}, \dots, B_{10}$ and five girls $G_{1}, G_{2}, \dots, G_{5}$ in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both $B_{1}$ and $B_{2}$ together should not be the members of a group, is ______.

Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is:

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:

The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to :

There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is :