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Trevor has 6 books on his summer reading list. He wants to choose 3 books to pack them in his suitcase for reading during vacation. How many different choices are possible? SHOW WORK

Trevor has 6 books on his summer reading list. He wants to choose 3 books to pack them in his suitcase for reading during vacation. How many different choices are possible? SHOW WORK

Algebra for College Students
Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter2: Equations, Inequalities, And Problem Solving
Section2.4: Formulas
Problem 64PS
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**Problem 11:** Trevor has 6 books on his summer reading list. He wants to choose 3 books to pack them in his suitcase for reading during vacation. How many different choices are possible? SHOW WORK **Solution:** To determine the number of ways Trevor can choose 3 books out of 6, we can use the concept of combinations. The formula for combinations 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 (the product of all positive integers up to that number). In this problem, \( n = 6 \) and \( k = 3 \). Plugging these values into the formula, we have: \[ C(6, 3) = \frac{6!}{3!(6-3)!} \] \[ C(6, 3) = \frac{6!}{3! \cdot 3!} \] \[ C(6, 3) = \frac{6 \times 5 \times 4 \times 3!}{3! \times 3!} \] \[ C(6, 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \] \[ C(6, 3) = \frac{120}{6} \] \[ C(6, 3) = 20 \] Therefore, Trevor has 20 different choices for selecting 3 books out of his 6-book reading list.
Transcribed Image Text:**Problem 11:** Trevor has 6 books on his summer reading list. He wants to choose 3 books to pack them in his suitcase for reading during vacation. How many different choices are possible? SHOW WORK **Solution:** To determine the number of ways Trevor can choose 3 books out of 6, we can use the concept of combinations. The formula for combinations 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 (the product of all positive integers up to that number). In this problem, \( n = 6 \) and \( k = 3 \). Plugging these values into the formula, we have: \[ C(6, 3) = \frac{6!}{3!(6-3)!} \] \[ C(6, 3) = \frac{6!}{3! \cdot 3!} \] \[ C(6, 3) = \frac{6 \times 5 \times 4 \times 3!}{3! \times 3!} \] \[ C(6, 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \] \[ C(6, 3) = \frac{120}{6} \] \[ C(6, 3) = 20 \] Therefore, Trevor has 20 different choices for selecting 3 books out of his 6-book reading list.
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