You are trying to decide what meal to serve for the student engagement award dinner. To help you decide you took a survey 30 students and asked them if they liked pasta chicken or steak. Three students don't like any of those options. 14 people like pasta. 15 people like chicken. 11 people like steak. Six people like pasta and chicken. Five people like pasta and steak. Four people like chicken and steak. Two people like all three.

This is a problem involving set theory, particularly Venn diagrams. Let’s break it down step by step, where:

- \( P \) represents students who like **pasta**.
- \( C \) represents students who like **chicken**.
- \( S \) represents students who like **steak**.

Given:
- Total students surveyed: 30
- Students who don't like any options: 3
- Students who like pasta: \( |P| = 14 \)
- Students who like chicken: \( |C| = 15 \)
- Students who like steak: \( |S| = 11 \)
- Students who like both pasta and chicken: \( |P \cap C| = 6 \)
- Students who like both pasta and steak: \( |P \cap S| = 5 \)
- Students who like both chicken and steak: \( |C \cap S| = 4 \)
- Students who like all three: \( |P \cap C \cap S| = 2 \)

### Steps:

#### 1. Calculate the number of students who like exactly all three:
- \( |P \cap C \cap S| = 2 \)

#### 2. Calculate the number of students who like exactly two out of the three options:

- For pasta and chicken only: \( |P \cap C| - |P \cap C \cap S| = 6 - 2 = 4 \)
- For pasta and steak only: \( |P \cap S| - |P \cap C \cap S| = 5 - 2 = 3 \)
- For chicken and steak only: \( |C \cap S| - |P \cap C \cap S| = 4 - 2 = 2 \)

#### 3. Calculate the number of students who like only one option:

- For pasta only: 
  \( |P| - \text{(students who like pasta and another)} = 14 - (4 + 3 + 2) = 14 - 9 = 5 \)
  
- For chicken only: 
  \( |C| - \text{(students who like chicken and another)} = 15 - (4 + 2 + 2) = 15 - 8 = 7 \)
  
- For steak only: 
  \( |S| - \text{(students who like steak and another)} = 11 - (3 + 2 + 2) = 11 - 7 = 4 \)

#### 4. Double-check total:
- Add all groups:
  \[
  5 \,(\text{pasta only}) + 7 \,(\text{chicken only}) + 4 \,(\text{steak only}) + 4 \,(\text{pasta and chicken only}) + 3 \,(\text{pasta and steak only}) + 2 \,(\text{chicken and steak only}) + 2 \,(\text{all three}) = 27
  \]
- Plus 3 students who don't like any options: \( 27 + 3 = 30 \), which matches the total number surveyed.

### Conclusion:
- 5 students like only pasta.
- 7 students like only chicken.
- 4 students like only steak.
- 4 students like both pasta and chicken.
- 3 students like both pasta and steak.
- 2 students like both chicken and steak.
- 2 students like all three.
- 3 students don't like any of the options.

If you want a more detailed explanation of any part, feel free to ask! 

**Related Questions:**
1. How many students like only one of the meal options?
2. How many students like more than one meal option?
3. What is the probability of randomly selecting a student who likes pasta?
4. What is the union of students who like pasta or chicken?
5. How many students like steak but not pasta?

**Tip:** Venn diagrams are a great visual tool to break down problems involving intersections of sets like this one!

You are trying to decide what meal to serve for the student engagement award dinner. To help you decide you took a survey 30 students and asked them if they liked pasta, chicken, or steak. Three students don't like any of those options. 14 people like pasta. 15 people like chicken. 11 people like steak. Six people like pasta and chicken. Five people like pasta and steak. Four people like chicken and steak. Two people like all three.

This problem involves using set theory to analyze a survey of 30 students who were asked if they liked pasta, chicken, or steak. Explore how to calculate the number of students who like only one option, two options, or all three, using the Inclusion-Exclusion Principle. Perfect for students in Grades 9-12.

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