7) Suppose that a class of 20 students is assigned to write an essay. Suppose 4 essays are picked at random. How many groups of 4 are possible? What if the 4 essays picked at random are given awards of $35, $20, $10, and $5. How many groups of 4 are possible now?
7) Suppose that a class of 20 students is assigned to write an essay. Suppose 4 essays are picked at random. How many groups of 4 are possible? What if the 4 essays picked at random are given awards of $35, $20, $10, and $5. How many groups of 4 are possible now?
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![Certainly! Here's the transcription suitable for an educational website:
---
**Suppose that a class of 20 students is assigned to write an essay.**
**Question:**
1. **Suppose 4 essays are picked at random.**
- How many groups of 4 are possible?
2. **What if the 4 essays picked at random are given awards of $35, $20, $10, and $5?**
- How many groups of 4 are possible now?
In this scenario, we want to explore two different types of combinations:
- **Combination without regard to order (unlabeled):** This is the case in which essays are selected purely based on the group, without any distinction between them.
- **Combination with regard to order (labeled):** This is the case where each essay picked is assigned a specific award, implying that the order of selection now matters.
---
### Explanation of Concepts:
1. **Unordered Combinations (Unlabeled):**
- This refers to selecting 4 essays from the 20 in the class without considering any order of selection.
- To calculate this, we use the combination formula \(\binom{n}{k}\), where \(n\) is the total number of items to choose from, and \(k\) is the number of items to choose.
- Formula: \(\binom{20}{4} = \frac{20!}{4!(20-4)!}\)
2. **Ordered Combinations (Labeled):**
- When the 4 essays are awarded specific amounts of money ($35, $20, $10, and $5), the order in which the essays are chosen now matters.
- This is a factorial problem represented by \(n!\), where each position is distinct.
- Formula: \(P(20, 4) = 20!/(20-4)!\)
### Calculation Illustrations:
- **Unordered Combinations:**
\[
\binom{20}{4} = \frac{20!}{4!(20-4)!} = \frac{20!}{4! \cdot 16!} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845
\]
Therefore, 4,845 different groups of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff6ad1cd7-e883-4155-acb4-5513247aaa08%2F35d0279e-dd26-43a2-a8a5-1100ec3f441a%2Fgs8gqrp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Certainly! Here's the transcription suitable for an educational website:
---
**Suppose that a class of 20 students is assigned to write an essay.**
**Question:**
1. **Suppose 4 essays are picked at random.**
- How many groups of 4 are possible?
2. **What if the 4 essays picked at random are given awards of $35, $20, $10, and $5?**
- How many groups of 4 are possible now?
In this scenario, we want to explore two different types of combinations:
- **Combination without regard to order (unlabeled):** This is the case in which essays are selected purely based on the group, without any distinction between them.
- **Combination with regard to order (labeled):** This is the case where each essay picked is assigned a specific award, implying that the order of selection now matters.
---
### Explanation of Concepts:
1. **Unordered Combinations (Unlabeled):**
- This refers to selecting 4 essays from the 20 in the class without considering any order of selection.
- To calculate this, we use the combination formula \(\binom{n}{k}\), where \(n\) is the total number of items to choose from, and \(k\) is the number of items to choose.
- Formula: \(\binom{20}{4} = \frac{20!}{4!(20-4)!}\)
2. **Ordered Combinations (Labeled):**
- When the 4 essays are awarded specific amounts of money ($35, $20, $10, and $5), the order in which the essays are chosen now matters.
- This is a factorial problem represented by \(n!\), where each position is distinct.
- Formula: \(P(20, 4) = 20!/(20-4)!\)
### Calculation Illustrations:
- **Unordered Combinations:**
\[
\binom{20}{4} = \frac{20!}{4!(20-4)!} = \frac{20!}{4! \cdot 16!} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845
\]
Therefore, 4,845 different groups of
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