Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ items from a set of $n$ distinct items without regard to order is given by $^nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$.
• Principle of Multiplication: If there are $m$ ways to do one thing and $n$ ways to do another, then there are $m \times n$ ways to do both.
• Principle of Addition: If there are $m$ ways to do one thing and $n$ ways to do another, and the two things cannot be done at the same time, then there are $m + n$ ways to do either one.

Step-by-Step Solution

Step 1: Determine Possible Distributions of Questions Across Sections

We need to find all possible combinations of questions chosen from the three sections (Section 1, Section 2, and Section 3) such that the total number of questions chosen is 5, and at least one question is chosen from each section. Let $x_1$, $x_2$, and $x_3$ represent the number of questions chosen from Section 1, Section 2, and Section 3, respectively. We have the equation:

$x_1 + x_2 + x_3 = 5$

with the constraints $x_1 \ge 1$, $x_2 \ge 1$, and $x_3 \ge 1$.

To simplify the problem, let $y_i = x_i - 1$ for $i = 1, 2, 3$. Then $y_i \ge 0$, and the equation becomes:

$(y_1 + 1) + (y_2 + 1) + (y_3 + 1) = 5$ $y_1 + y_2 + y_3 = 5 - 3 = 2$

Now we need to find the non-negative integer solutions to $y_1 + y_2 + y_3 = 2$. The possible solutions for $(y_1, y_2, y_3)$ are:

• (2, 0, 0), (0, 2, 0), (0, 0, 2)
• (1, 1, 0), (1, 0, 1), (0, 1, 1)

Converting back to $x_i$ values using $x_i = y_i + 1$, we get the following solutions for $(x_1, x_2, x_3)$:

• (3, 1, 1), (1, 3, 1), (1, 1, 3) corresponding to (2,0,0), (0,2,0), (0,0,2)
• (2, 2, 1), (2, 1, 2), (1, 2, 2) corresponding to (1,1,0), (1,0,1), (0,1,1)

Thus, the possible distributions of questions are (3, 1, 1) and (2, 2, 1).

Step 2: Calculate Ways for Each Distribution Pattern

We consider each case separately:

Case 1: The distribution of questions is (3, 1, 1)

This means one section contributes 3 questions, and the other two sections contribute 1 question each. There are 3 ways to choose which section contributes 3 questions. For each of these arrangements, we choose 3 questions from that section (in $^5C_3$ ways) and 1 question from each of the other two sections (in $^5C_1$ ways each).

The number of ways for this case is: $3 \times ^5C_3 \times ^5C_1 \times ^5C_1 = 3 \times \frac{5!}{3!2!} \times \frac{5!}{1!4!} \times \frac{5!}{1!4!} = 3 \times 10 \times 5 \times 5 = 750$

Case 2: The distribution of questions is (2, 2, 1)

This means two sections contribute 2 questions each, and the third section contributes 1 question. There are 3 ways to choose which section contributes 1 question. For each of these arrangements, we choose 2 questions from the first of the remaining sections (in $^5C_2$ ways), 2 questions from the second of the remaining sections (in $^5C_2$ ways), and 1 question from the last section (in $^5C_1$ ways).

The number of ways for this case is: $3 \times ^5C_2 \times ^5C_2 \times ^5C_1 = 3 \times \frac{5!}{2!3!} \times \frac{5!}{2!3!} \times \frac{5!}{1!4!} = 3 \times 10 \times 10 \times 5 = 1500$

Step 3: Calculate Total Number of Ways

Since the two cases are mutually exclusive, we add the number of ways for each case to find the total number of ways to choose the questions:

Total ways = Ways for Case 1 + Ways for Case 2 = $750 + 1500 = 2250$

Common Mistakes & Tips

• Forgetting the constraint: Always remember the "at least one question from each section" constraint.
• Overcounting: Ensure that you are not overcounting any combinations. Breaking the problem into mutually exclusive cases helps prevent this.
• Incorrectly applying combinations: Double-check your combination calculations.

Summary

We first determined the possible distributions of questions across the three sections, ensuring that at least one question was chosen from each section. We identified two possible distributions: (3, 1, 1) and (2, 2, 1). We then calculated the number of ways to choose questions for each distribution and summed the results to obtain the total number of ways. The total number of ways is 2250.

Final Answer

The final answer is $\boxed{2250}$, which corresponds to option (A).