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

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