In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper

Q: Why can't we just use the 'Stars and Bars' method here?

That is a fantastic question! 'Stars and Bars' is a powerful tool for distributing identical items into distinct bins. However, here, the questions are distinct (we are choosing specific questions, not just counting them), and we have upper-bound constraints (no more than 6 in B or C). Using 'Stars and Bars' would require inclusion-exclusion, which is far more complex than the direct case-by-case method we used.

Q: Wait, why did we reject $x = 8$ immediately?

Look at the constraints. If $x = 8$, then $y + z = 15 - 8 = 7$. But we know that $y \ge 4$ and $z \ge 4$. Therefore, the minimum value for $y + z$ is $4 + 4 = 8$. Since $7 < 8$, it is physically impossible to satisfy the minimum requirements if we pick 8 questions from Section A. Always check your boundary conditions first!

Q: Is there a faster way to solve this without listing every case?

You could use generating functions, where the answer is the coefficient of $x^{15}$ in the expansion of $(x^4 + \dots + x^8)(x^4 + \dots + x^6)^2$. However, for a JEE Advanced problem with such small numbers, the case-by-case method is actually faster and significantly less prone to algebraic errors. Never underestimate the power of a well-organized manual count!

Question

In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections: A, B and C. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section A has 8 questions, section B has 6 questions and section C has 6 questions, then the total number of ways a student can select 15 questions is_________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To solve this combinatorial problem, we define $x$, $y$, and $z$ as the number of questions selected from sections A, B, and C, respectively. The total number of questions to be selected is given by the equation:

$$x + y + z = 15$$

We are constrained by the number of available questions in each section, leading to the following inequalities:

$$4 \le x \le 8, \quad 4 \le y \le 6, \quad 4 \le z \le 6$$

### The Systematic Siege

We begin by testing the upper limits of our variables. If $x = 8$, then $y + z = 7$. However, since $y \ge 4$ and $z \ge 4$, the minimum sum for $y + z$ is $8$. Because $7 < 8$, the case $x = 8$ is impossible.

Next, we consider $x = 7$. This implies $y + z = 8$. Given the constraints, the only valid solution is $(y, z) = (4, 4)$. The number of ways is:

$${}^{8}C_{7} 	imes {}^{6}C_{4} 	imes {}^{6}C_{4} = 8 	imes 15 	imes 15 = 1800$$

### The Beauty of Symmetry

When $x = 6$, we require $y + z = 9$. The possible pairs for $(y, z)$ are $(5, 4)$ and $(4, 5)$.

For the case $(6, 5, 4)$, the number of ways is:

$${}^{8}C_{6} 	imes {}^{6}C_{5} 	imes {}^{6}C_{4} = 28 	imes 6 	imes 15 = 2520$$

Due to symmetry, the case $(6, 4, 5)$ yields the same result. Thus, we have $2 	imes 2520 = 5040$ ways for $x = 6$.

Moving to $x = 5$, we require $y + z = 10$. This yields three sub-cases: $(5, 6, 4)$, $(5, 4, 6)$, and $(5, 5, 5)$.

The extreme cases $(5, 6, 4)$ and $(5, 4, 6)$ each provide:

$${}^{8}C_{5} 	imes {}^{6}C_{6} 	imes {}^{6}C_{4} = 56 	imes 1 	imes 15 = 840$$

Totaling $840 	imes 2 = 1680$ ways. The balanced case $(5, 5, 5)$ provides:

$${}^{8}C_{5} 	imes {}^{6}C_{5} 	imes {}^{6}C_{5} = 56 	imes 6 	imes 6 = 2016$$

Finally, for $x = 4$, we require $y + z = 11$. The only valid pairs are $(6, 5)$ and $(5, 6)$. Each provides:

$${}^{8}C_{4} 	imes {}^{6}C_{6} 	imes {}^{6}C_{5} = 70 	imes 1 	imes 6 = 420$$

Totaling $420 	imes 2 = 840$ ways.

### Final Calculation

We now aggregate the results from all valid configurations to find the total number of ways:

$$1800 + 5040 + 1680 + 2016 + 840 = 11376$$

The total number of ways to select the questions is **11376**.

Enter Numerical Value:

Visualized Solution

Understanding the Paper Structure

Defining Mathematical Constraints

Case 1: Maximum from Section A

Case 2 & 3: Selecting 6 from Section A

Case 4 & 5: Selecting 5 from Section A (Part 1)

Case 6: Selecting 5 from Section A (Part 2)

Case 7 & 8: Minimum from Section A

Summing All Possibilities

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Systematic Siege

The Beauty of Symmetry

Final Calculation

Similar Questions

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 :

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

A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is

The number of ways of getting a sum 16 on throwing a dice four times is______

The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is _____

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 ____.

There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2 are wicketkeepers. The number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsmen and 1 wicketkeeper, is ___

Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to

The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is ______.