Can someone help with number 7? 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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Description: Can someone help with number 7? 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 g

Total number of students in class = 20 Number of essays to be picked at random = 4

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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?

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Can someone help with number 7?

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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