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1. How many ways can you choose 4 numbers from the set p = {4, 6, 8, 10, 12} or 3 numbers from the set q = {11, 12, 13, 14}? Show your work in the space provided.

1. How many ways can you choose 4 numbers from the set p = {4, 6, 8, 10, 12} or 3 numbers from the set q = {11, 12, 13, 14}? Show your work in the space provided.

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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### Problem Description 1. **Combinatorial Question**: How many ways can you choose 4 numbers from the set \( p = \{4, 6, 8, 10, 12\} \) or 3 numbers from the set \( q = \{11, 12, 13, 14\} \)? - **Task**: Show your work in the space provided. ### Explanation of the Problem: In this problem, you are asked to calculate the number of combinations for two different scenarios involving two different sets of numbers. 1. **Choosing 4 numbers from Set \( p \)**: - Set \( p \) is given as \( \{4, 6, 8, 10, 12\} \), which contains 5 elements. - You need to find the number of ways to choose 4 elements from this 5-element set. 2. **Choosing 3 numbers from Set \( q \)**: - Set \( q \) is given as \( \{11, 12, 13, 14\} \), which contains 4 elements. - You need to find the number of ways to choose 3 elements from this 4-element set. ### Solution Process: For each scenario, you will use the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of elements, and \( k \) is the number of elements to choose. 1. **For Set \( p \)**: - \( n = 5 \) (since the set has 5 elements) - \( k = 4 \) (since you need to choose 4 elements) - Calculate \( C(5, 4) \) 2. **For Set \( q \)**: - \( n = 4 \) (since the set has 4 elements) - \( k = 3 \) (since you need to choose 3 elements) - Calculate \( C(4, 3) \) ### Calculation: 1. **For Set \( p \) (Choosing 4 out of 5 elements)**: \[ C(5, 4) = \frac{5!}{4!(5-4)!} = \frac{5!}{4! \cdot
Transcribed Image Text:### Problem Description 1. **Combinatorial Question**: How many ways can you choose 4 numbers from the set \( p = \{4, 6, 8, 10, 12\} \) or 3 numbers from the set \( q = \{11, 12, 13, 14\} \)? - **Task**: Show your work in the space provided. ### Explanation of the Problem: In this problem, you are asked to calculate the number of combinations for two different scenarios involving two different sets of numbers. 1. **Choosing 4 numbers from Set \( p \)**: - Set \( p \) is given as \( \{4, 6, 8, 10, 12\} \), which contains 5 elements. - You need to find the number of ways to choose 4 elements from this 5-element set. 2. **Choosing 3 numbers from Set \( q \)**: - Set \( q \) is given as \( \{11, 12, 13, 14\} \), which contains 4 elements. - You need to find the number of ways to choose 3 elements from this 4-element set. ### Solution Process: For each scenario, you will use the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of elements, and \( k \) is the number of elements to choose. 1. **For Set \( p \)**: - \( n = 5 \) (since the set has 5 elements) - \( k = 4 \) (since you need to choose 4 elements) - Calculate \( C(5, 4) \) 2. **For Set \( q \)**: - \( n = 4 \) (since the set has 4 elements) - \( k = 3 \) (since you need to choose 3 elements) - Calculate \( C(4, 3) \) ### Calculation: 1. **For Set \( p \) (Choosing 4 out of 5 elements)**: \[ C(5, 4) = \frac{5!}{4!(5-4)!} = \frac{5!}{4! \cdot
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