ASK YOUR TEACHER A student is given a reading list of nine books from which he must select three for an outside reading requirement. In how many ways can he make his selections? ways Need Help? Read It Show My Work (Optional)? PRACTICE ANOTHER
ASK YOUR TEACHER A student is given a reading list of nine books from which he must select three for an outside reading requirement. In how many ways can he make his selections? ways Need Help? Read It Show My Work (Optional)? PRACTICE ANOTHER
Chapter8: Sequences, Series,and Probability
Section8.6: Counting Principles
Problem 17E
Related questions
Question
![### Combinatorics Problem - Book Selection
**Problem Statement:**
A student is given a reading list of nine books from which he must select three for an outside reading requirement. In how many ways can he make his selections?
---
**Explanation:**
This is a classic problem of combinations in combinatorics, where the order of selection does not matter. The solution involves calculating the combination of 9 books taken 3 at a time, which is denoted mathematically as \( \binom{9}{3} \).
The formula for combinations is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Where:
- \( n \) is the total number of items (books),
- \( r \) is the number of items to choose (books to be selected),
- \( ! \) denotes factorial, the product of all positive integers up to that number.
For this problem:
- \( n = 9 \),
- \( r = 3 \).
Substitute the values into the formula:
\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} \].
Calculate the factorials:
- \( 9! = 9 \times 8 \times 7 \times 6! \).
Thus:
\[ \binom{9}{3} = \frac{9 \times 8 \times 7 \times 6!}{3! \times 6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84 \].
**Conclusion:**
The number of ways the student can select three books from a list of nine is 84.
---
**Interactive Elements:**
- **Need Help?** - Button for additional hints or guidance.
- **Show My Work (Optional)** - Option for students to show their solution process.
- **Submit Answer** - Button for students to submit their answers.
Additional interactive elements include navigation buttons for notes, teacher assistance, and practice problems.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e1b5fc2-7275-49e6-a434-e3b12d2d745f%2F76a767de-4861-4ab7-b6d0-336272926d05%2Fq6xybi7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Combinatorics Problem - Book Selection
**Problem Statement:**
A student is given a reading list of nine books from which he must select three for an outside reading requirement. In how many ways can he make his selections?
---
**Explanation:**
This is a classic problem of combinations in combinatorics, where the order of selection does not matter. The solution involves calculating the combination of 9 books taken 3 at a time, which is denoted mathematically as \( \binom{9}{3} \).
The formula for combinations is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Where:
- \( n \) is the total number of items (books),
- \( r \) is the number of items to choose (books to be selected),
- \( ! \) denotes factorial, the product of all positive integers up to that number.
For this problem:
- \( n = 9 \),
- \( r = 3 \).
Substitute the values into the formula:
\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} \].
Calculate the factorials:
- \( 9! = 9 \times 8 \times 7 \times 6! \).
Thus:
\[ \binom{9}{3} = \frac{9 \times 8 \times 7 \times 6!}{3! \times 6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84 \].
**Conclusion:**
The number of ways the student can select three books from a list of nine is 84.
---
**Interactive Elements:**
- **Need Help?** - Button for additional hints or guidance.
- **Show My Work (Optional)** - Option for students to show their solution process.
- **Submit Answer** - Button for students to submit their answers.
Additional interactive elements include navigation buttons for notes, teacher assistance, and practice problems.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 1 images
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Holt Mcdougal Larson Pre-algebra: Student Edition…
Algebra
ISBN:
9780547587776
Author:
HOLT MCDOUGAL
Publisher:
HOLT MCDOUGAL
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Holt Mcdougal Larson Pre-algebra: Student Edition…
Algebra
ISBN:
9780547587776
Author:
HOLT MCDOUGAL
Publisher:
HOLT MCDOUGAL
Algebra and Trigonometry (MindTap Course List)
Algebra
ISBN:
9781305071742
Author:
James Stewart, Lothar Redlin, Saleem Watson
Publisher:
Cengage Learning
Glencoe Algebra 1, Student Edition, 9780079039897…
Algebra
ISBN:
9780079039897
Author:
Carter
Publisher:
McGraw Hill
