Question

A student must answer 8 out of 10 questions on an exam. (a) How many choices does the student have? (b) How many choices does a student have if the first three questions must be answered? (c) How many choices does a student have if exactly four out of the first five questions must be answered?

Short answer

(a) 45 choices. (b) 21 choices. (c) 25 choices.

Step-by-step solution

Understanding the Problem

The exercise involves combinatorial choices, which include selecting a subset of questions from a given set.

Calculate Choices for (a)

(a) The student needs to choose 8 questions out of the total 10. The formula for combinations is used here: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] Calculate \( \binom{10}{8} \), which simplifies to \( \binom{10}{2} \), yielding a result of \(45\) choices.

Calculate Choices for (b)

(b) The first 3 questions must be included, so the student needs to choose 5 more questions out of the remaining 7 questions. Use the combination formula: \[ \binom{7}{5} = \binom{7}{2} = 21 \] Thus, there are 21 ways to choose the remaining questions.

Calculate Choices for (c)

(c) The student must choose exactly 4 out of the first 5 questions. For the remaining 4 questions, the student needs to select 4 more from the remaining 5 questions (after choosing 4 of the initial ones). Calculate: \[ \binom{5}{4} \times \binom{5}{4} = 5 \times 5 = 25 \] Therefore, the student has 25 choices.

Key concepts

Understanding Combinations

Combinations are a key concept in combinatorics, a branch of mathematics focused on counting, arranging, and finding patterns within particular sets. When dealing with combinations, order does not matter. This means selecting three apples from a basket results in the same set whether you pick them the left, right, or center first.

The formula used to calculate combinations is represented by \( \binom{n}{r} \), which refers to "\( n \) choose \( r \)". It is expressed mathematically as:

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

Here, \( n \) indicates the total number of items to choose from, and \( r \) represents the number of selections made. For example, picking 8 questions out of 10 on an exam, you'd compute \( \binom{10}{8} \), which simplifies to \( \binom{10}{2} \) as choosing 8 questions is the same as not choosing 2 questions. Thus, the result is 45 possible combinations.

Breaking Down Problem-Solving Steps

Problem-solving in combinatorics requires a methodical approach. Let's walk through the steps to tackle these types of problems effectively.

Start by understanding the problem's requirements. In this case, we need to determine how many ways a student can answer certain questions on an exam. Once the problem is understood, identify the type of combinatorial structure needed, which is often either a combination or a permutation.

Next, perform the necessary calculations using the correct formulas. As shown in the exercise, it's crucial to adapt the formula based on given conditions, such as mandatory questions or limited choices.

Finally, evaluate your results for accuracy and practicality, ensuring all stated requirements are met. By following these steps, tackling exam-related combinations becomes systematic and manageable.

Optimizing for Exam Question Selection

Selecting questions during an exam can be strategic, especially when choices need to be made from a set number of possible questions. This involves analyzing the combination scenarios to maximize your score or meet specific criteria.

For instance, consider a scenario where a student has to answer a subset of questions where some are mandatory. Here, knowing how to effectively apply the combinations formula helps determine how the rest of the questions can be chosen.

• In scenario (a), the student freely chooses any 8 out of 10 questions.
• In scenario (b), 3 questions are fixed, and the student selects the remaining from the rest.
• In scenario (c), choices depend on answering a set number from the first part of the test.

Mastering these selection methods allows students to confidently approach their tests or any similar selection-based problems.

Understanding Permutations

Permutations differ from combinations because the order does matter. Permutations are used when every distinct arrangement of a set influences the outcome. Think of it like rearranging letters in a word where every different lineup forms a new permutation.

The calculations for permutations look slightly different as well, using the formula \( P(n, r) = \frac{n!}{(n-r)!} \), which finds the number of ways to arrange \( r \) items from \( n \).

• This principle applies when choices or sequences need specific ordering.
• Unlike combinations, a permutation will vary its result if the sequence changes even slightly.

In exams, while permutations are less about selecting questions, they are crucial in subjects where sequences or series of actions are pivotal. Understanding when to apply permutations over combinations equips students with the right tool for every combinatorial problem.