Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ objects from a set of $n$ distinct objects, without regard to order, is given by the binomial coefficient: $${n \choose r} = \frac{n!}{r!(n-r)!}$$
• Inclusion-Exclusion Principle (simplified): When counting the number of ways to do something with multiple conditions, we often need to consider all possible cases that satisfy the given constraints.

Step-by-Step Solution

Step 1: Identify the constraints

A student must choose 15 questions from 20, with at least 4 from each section. Section A has 8 questions, and sections B and C have 6 each.

Step 2: Enumerate all possible combinations of question selections from each section that satisfy the constraints.

Let $a$, $b$, and $c$ be the number of questions chosen from sections A, B, and C, respectively. We need to find all non-negative integer solutions to the equation $a + b + c = 15$, subject to the constraints:

• $4 \le a \le 8$
• $4 \le b \le 6$
• $4 \le c \le 6$

We can list all such combinations:

1. $a=4, b=5, c=6$
2. $a=4, b=6, c=5$
3. $a=5, b=4, c=6$
4. $a=5, b=5, c=5$
5. $a=5, b=6, c=4$
6. $a=6, b=4, c=5$
7. $a=6, b=5, c=4$
8. $a=7, b=4, c=4$

Step 3: Calculate the number of ways to choose questions for each combination

For each combination $(a, b, c)$, the number of ways to choose the questions is given by ${8 \choose a} \cdot {6 \choose b} \cdot {6 \choose c}$. We calculate this for each case:

1. $a=4, b=5, c=6$: ${8 \choose 4} {6 \choose 5} {6 \choose 6} = \frac{8!}{4!4!} \cdot \frac{6!}{5!1!} \cdot \frac{6!}{6!0!} = 70 \cdot 6 \cdot 1 = 420$
2. $a=4, b=6, c=5$: ${8 \choose 4} {6 \choose 6} {6 \choose 5} = \frac{8!}{4!4!} \cdot \frac{6!}{6!0!} \cdot \frac{6!}{5!1!} = 70 \cdot 1 \cdot 6 = 420$
3. $a=5, b=4, c=6$: ${8 \choose 5} {6 \choose 4} {6 \choose 6} = \frac{8!}{5!3!} \cdot \frac{6!}{4!2!} \cdot \frac{6!}{6!0!} = 56 \cdot 15 \cdot 1 = 840$
4. $a=5, b=5, c=5$: ${8 \choose 5} {6 \choose 5} {6 \choose 5} = \frac{8!}{5!3!} \cdot \frac{6!}{5!1!} \cdot \frac{6!}{5!1!} = 56 \cdot 6 \cdot 6 = 2016$
5. $a=5, b=6, c=4$: ${8 \choose 5} {6 \choose 6} {6 \choose 4} = \frac{8!}{5!3!} \cdot \frac{6!}{6!0!} \cdot \frac{6!}{4!2!} = 56 \cdot 1 \cdot 15 = 840$
6. $a=6, b=4, c=5$: ${8 \choose 6} {6 \choose 4} {6 \choose 5} = \frac{8!}{6!2!} \cdot \frac{6!}{4!2!} \cdot \frac{6!}{5!1!} = 28 \cdot 15 \cdot 6 = 2520$
7. $a=6, b=5, c=4$: ${8 \choose 6} {6 \choose 5} {6 \choose 4} = \frac{8!}{6!2!} \cdot \frac{6!}{5!1!} \cdot \frac{6!}{4!2!} = 28 \cdot 6 \cdot 15 = 2520$
8. $a=7, b=4, c=4$: ${8 \choose 7} {6 \choose 4} {6 \choose 4} = \frac{8!}{7!1!} \cdot \frac{6!}{4!2!} \cdot \frac{6!}{4!2!} = 8 \cdot 15 \cdot 15 = 1800$

Step 4: Sum the results

The total number of ways is the sum of the ways for each combination: $420 + 420 + 840 + 2016 + 840 + 2520 + 2520 + 1800 = 11376$

Common Mistakes & Tips

• Forgetting Constraints: Always double-check that each combination you consider satisfies all the given constraints (minimum number of questions from each section, and total number of questions).
• Double Counting: Ensure that you are not counting any combinations more than once. Listing them out systematically helps.
• Calculation Errors: Be careful with the factorials when calculating the binomial coefficients.

Summary

We enumerated all possible combinations of questions from each section that satisfied the given constraints, calculated the number of ways to choose questions for each combination using binomial coefficients, and then summed the results to find the total number of ways a student can select the questions.

The final answer is \boxed{11376}.