Problem 41 - Algebra 2

Question

REVIEW An examination consists of 10 questions. A student must answer only one of the first two questions and only six of the remaining ones. How many choices of questions does the student have? \(\mathrm{F} 112\) \(\mathrm{G} 56\) \(\mathrm{H} 44\) \(\mathrm{J} 30\)

Step-by-Step Solution

Answer

The student has 56 choices of questions.

Step 1: Choose from the First Two Questions

The student must choose only one of the first two questions. There are 2 choices here: either the first question or the second question can be chosen.

Step 2: Choose Six of the Remaining Questions

After answering one of the first two questions, the student has 8 questions left to choose from. From these, the student needs to choose 6. The number of ways to choose 6 questions from 8 is given by the binomial coefficient \( \binom{8}{6} \).

Step 3: Calculate Combinations

Calculate \( \binom{8}{6} \), which is equivalent to calculating \( \binom{8}{2} \) because \( \binom{n}{k} = \binom{n}{n-k} \). \( \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \).

Step 4: Combine Choices

Since there are 2 options for the first question and 28 combinations for selecting the remaining 6 questions, multiply these results: \( 2 \times 28 = 56 \). This is the total number of ways the student can choose the questions.

Key Concepts

Binomial Coefficient

The binomial coefficient is a fundamental concept in combinatorics. It represents the number of ways to choose a subset of "k" elements from a larger set of "n" elements without considering the order. It is denoted as \( \binom{n}{k} \). This coefficient helps us solve problems involving combinations, like how many different teams you can form from a group of people.

• Formula: The binomial coefficient is calculated using the formula: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) where "!" denotes factorial, meaning the product of all positive integers up to that number.
• Applications: It is widely used in probability, statistics, and algebraic expressions expansion, like in the binomial theorem.

In the given exam exercise, we used a binomial coefficient to choose 6 questions from the remaining 8 questions after one of the first two questions was answered.

Combinations

Combinations are selections of items from a larger pool where the order of selection does not matter. This is different from permutations, where order is important.

• When calculating combinations, we use the binomial coefficient: \( \binom{n}{k} \).
• Example: For the exam question, the student chooses 6 out of 8 remaining questions. So, we calculate \( \binom{8}{6} \).

After choosing one of the first two questions (2 ways), the student calculates the combinations of the next choices as a simple product of combinations (first question selection) with other possible selections.

Permutations

Permutations refer to arrangements where the order does matter. This is different from combinations, where order is irrelevant.

• Formula: To find permutations of "n" items taken "k" at a time, use the formula: \( P(n, k) = \frac{n!}{(n-k)!} \).
• Key Difference: Since order matters in permutations, they often count more possible arrangements than combinations.

In our exercise, permutations were not directly involved because the order of answering questions does not change the selection process outcome.

Decision-Making in Exams

Decision-making in exams involves choosing questions tactically, especially when there are limited options you can select from, as shown in the given exercise about selecting questions.

• Strategic Thinking: First, decide the topic or type of question you are most confident about from the first two options.
• Combination Strategy: For the rest, calculate how many combinations fit your strength and what you feel comfortable tackling.

Using combinatorics, students can determine the number of potential ways they can select questions, increasing preparedness and confidence on test day.