Problem 70 - College Algebra with Modeling and Visualization

Question

On a test with 6 essay questions, students are asked to answer 4 questions. How many ways can the essay questions be selected? in the second part a student must choose 4 of 5 essay questions. How many ways can the essay questions be selected?

Step-by-Step Solution

Answer

15 ways for the first part; 5 ways for the second part.

Step 1: Understanding the Problem

We need to determine the number of ways to select questions from a set. The first part has 6 questions and we need to choose 4. The second part has 5 questions and we also need to choose 4.

Step 2: Applying the Combination Formula

The combination formula is given by \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of items, \( k \) is the number of items to choose, and \( ! \) denotes factorial.

Step 3: Calculating for First Part: Selecting 4 out of 6

For the first part, we have \( n = 6 \) and \( k = 4 \). Using the formula: \[ C(6, 4) = \frac{6!}{4! \times (6-4)!} \]Calculate:\( 6! = 720 \), \( 4! = 24 \), \( 2! = 2 \). Therefore, \[ C(6, 4) = \frac{720}{24 \times 2} = \frac{720}{48} = 15 \].

Step 4: Calculating for Second Part: Selecting 4 out of 5

For the second part, we have \( n = 5 \) and \( k = 4 \). Using the formula: \[ C(5, 4) = \frac{5!}{4! \times (5-4)!} \]Calculate:\( 5! = 120 \), \( 4! = 24 \), \( 1! = 1 \). Therefore, \[ C(5, 4) = \frac{120}{24 \times 1} = \frac{120}{24} = 5 \].

Step 5: Final Answer

The number of ways to select questions for the first part is 15, and for the second part is 5.

Key Concepts

Combinations

Combinations are a fundamental concept in college algebra, especially when dealing with problems where order doesn't matter. In situations like choosing essay questions from a test, we're often interested in how many ways we can select a group from a larger set.
To do this, we use the combination formula. This helps us calculate the number of possible groups (or combinations) by taking into account the total number of items and the number of items we want to choose.

• The formula for combinations is given by: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where:
  • n is the total number of items, and
  • k is the number of items to choose.

The key aspect of combinations is that the order of selection doesn't matter, distinguishing them from permutations. For example, when selecting 4 questions out of 6, the group is the same regardless of the order the questions are picked.

Factorial

The term 'factorial' is important not just in combinations, but in many areas of mathematics. A factorial, denoted by an exclamation point (!), represents the product of all positive integers up to a specified number. For any positive integer \( n \), the factorial is the product:

• \[ n! = n \times (n-1) \times (n-2) \times \cdots \times 1 \]

For example, \( 4! \) (read as "four factorial") is:

• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)

Factorials are used in the combination formula to calculate the total number of possible arrangements or selections of items. For the exercise: when choosing 4 out of 6 questions, we use factorials to find the value of \( n! \), \( k! \), and \((n-k)!\). This allows us to find the number of combinations effectively.
Understanding factorial basics will hugely benefit your grasp of more complex mathematical problems involving permutations and combinations.

Counting Methods

Counting methods are essential techniques that help us tally the number of ways certain events can occur. In mathematics, effective counting is required for problems involving selections. Especially when using combinations, these methods give us a systematic way of finding the number of possible outcomes.
When dealing with algebraic problems like selecting essays, the combination formula is a key counting method. Here's how it's applied:

• Identify the total number of items (essays in this case). This is denoted as \( n \).
• Select the number of items you want to choose. This is \( k \).
• Apply the formula to obtain the number of combinations.
• Remember, the results will give the total number of ways to choose \( k \) items without regard to order.

By mastering counting methods like combinations, students can tackle a variety of problems with confidence, knowing that they're using a logical and systematic approach to finding solutions.