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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### 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?
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**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.
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