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 _______ .

Answer 1

Correct Answer: 11376

Define the variables: Let \( x_A \) be the number of questions selected from section A. Let \( x_B \) be the number of questions selected from section B. Let \( x_C \) be the number of questions selected from section C. We know that \( x_A + x_B + x_C = 15 \) with the constraints \( x_A \geq 4 \), \( x_B \geq 4 \), and \( x_C \geq 4 \).

Transform the variables: Introduce new variables:

\[ y_A = x_A - 4, \quad y_B = x_B - 4, \quad y_C = x_C - 4 \]

The equation becomes:

\[ y_A + y_B + y_C = 3 \]

Count the non-negative integer solutions: The number of non-negative integer solutions is given by:

\[ \text{Number of solutions} = \binom{n + k - 1}{k - 1} \]

Here, \( n = 3 \) and \( k = 3 \):

\[ \text{Number of solutions} = \binom{5}{2} = 10 \]

Calculate the total combinations: Total selections can be computed as:

\[ \text{Total ways} = \sum C(8, x_A) \times C(6, x_B) \times C(6, x_C) \]

\[ = 56 \times 6 + 28 \times 6 \times 15 \times 2 + 56 \times 15 \times 2 + 70 \times 6 \times 2 + 8 \times 15 \times 15 \]

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

Thus, the answer is: 11376

Answer 2

The problem asks for the total number of ways a student can select 15 questions from three sections A, B, and C, with specific constraints on the minimum number of questions to be attempted from each section.

Concept Used:

This problem is based on the concept of Combinations. The number of ways to select \( r \) items from a set of \( n \) distinct items is given by the formula:

\[ \binom{n}{r} = C(n, r) = \frac{n!}{r!(n-r)!} \]

Since the selections from different sections are independent events, we will use the Multiplication Principle to find the number of ways for each case, and the Addition Principle to find the total number of ways by summing up all possible cases.

Step-by-Step Solution:

Step 1: Define the variables and constraints.

Let \(x_A\), \(x_B\), and \(x_C\) be the number of questions selected from Section A, Section B, and Section C, respectively.

The number of available questions in each section are:

• Section A: 8 questions
• Section B: 6 questions
• Section C: 6 questions

The constraints are:

1. Total questions to be attempted: \( x_A + x_B + x_C = 15 \)
2. At least 4 questions from each section: \( x_A \geq 4, x_B \geq 4, x_C \geq 4 \)
3. The number of selected questions cannot exceed the available questions: \( x_A \leq 8, x_B \leq 6, x_C \leq 6 \)

Step 2: Identify all possible combinations (cases) of \((x_A, x_B, x_C)\) that satisfy the given constraints.

We need to find integer solutions to \( x_A + x_B + x_C = 15 \) under the conditions \(4 \le x_A \le 8\), \(4 \le x_B \le 6\), and \(4 \le x_C \le 6\). We can systematically list the cases by iterating through the possible values for \(x_B\).

• Case I: If \(x_B = 4\), then \(x_A + x_C = 11\). The possible pairs \((x_A, x_C)\) satisfying their constraints are:
  • \((7, 4)\) since \(x_A=7 \le 8\) and \(x_C=4 \le 6\). This gives the combination (7, 4, 4).
  • \((6, 5)\) since \(x_A=6 \le 8\) and \(x_C=5 \le 6\). This gives the combination (6, 4, 5).
  • \((5, 6)\) since \(x_A=5 \le 8\) and \(x_C=6 \le 6\). This gives the combination (5, 4, 6).
• Case II: If \(x_B = 5\), then \(x_A + x_C = 10\). The possible pairs \((x_A, x_C)\) are:
  • \((6, 4)\) since \(x_A=6 \le 8\) and \(x_C=4 \le 6\). This gives the combination (6, 5, 4).
  • \((5, 5)\) since \(x_A=5 \le 8\) and \(x_C=5 \le 6\). This gives the combination (5, 5, 5).
  • \((4, 6)\) since \(x_A=4 \le 8\) and \(x_C=6 \le 6\). This gives the combination (4, 5, 6).
• Case III: If \(x_B = 6\), then \(x_A + x_C = 9\). The possible pairs \((x_A, x_C)\) are:
  • \((5, 4)\) since \(x_A=5 \le 8\) and \(x_C=4 \le 6\). This gives the combination (5, 6, 4).
  • \((4, 5)\) since \(x_A=4 \le 8\) and \(x_C=5 \le 6\). This gives the combination (4, 6, 5).

Step 3: Calculate the number of ways for each valid case.

The number of ways for a selection \((x_A, x_B, x_C)\) is given by \( \binom{8}{x_A} \times \binom{6}{x_B} \times \binom{6}{x_C} \).

• For (7, 4, 4): \( \binom{8}{7} \times \binom{6}{4} \times \binom{6}{4} = 8 \times 15 \times 15 = 1800 \)
• For (6, 4, 5): \( \binom{8}{6} \times \binom{6}{4} \times \binom{6}{5} = 28 \times 15 \times 6 = 2520 \)
• For (5, 4, 6): \( \binom{8}{5} \times \binom{6}{4} \times \binom{6}{6} = 56 \times 15 \times 1 = 840 \)
• For (6, 5, 4): \( \binom{8}{6} \times \binom{6}{5} \times \binom{6}{4} = 28 \times 6 \times 15 = 2520 \)
• For (5, 5, 5): \( \binom{8}{5} \times \binom{6}{5} \times \binom{6}{5} = 56 \times 6 \times 6 = 2016 \)
• For (4, 5, 6): \( \binom{8}{4} \times \binom{6}{5} \times \binom{6}{6} = 70 \times 6 \times 1 = 420 \)
• For (5, 6, 4): \( \binom{8}{5} \times \binom{6}{6} \times \binom{6}{4} = 56 \times 1 \times 15 = 840 \)
• For (4, 6, 5): \( \binom{8}{4} \times \binom{6}{6} \times \binom{6}{5} = 70 \times 1 \times 6 = 420 \)

Final Computation & Result:

The total number of ways is the sum of the ways for all possible cases.

\[ \text{Total Ways} = 1800 + 2520 + 840 + 2520 + 2016 + 420 + 840 + 420 \] \[ \text{Total Ways} = 11376 \]

Thus, the total number of ways a student can select 15 questions is 11376.