Let's solve this problem step by step using the principle of inclusion-exclusion.

Definitions:

• Let $B$ represent the number of students with blond hair.
• Let $E$ represent the number of students with blue eyes.
• Let $n(B \cap E)$ represent the number of students with both blond hair and blue eyes.
• Let $n(B \cup E)$ represent the number of students with either blond hair or blue eyes or both.

Given:

• Total number of students: 45
• Number of students with both blond hair and blue eyes: $n(B \cap E) = 8$
• Number of students with neither blond hair nor blue eyes: 8
• Number of students with blond hair is twice the number of students with blue eyes: $B = 2E$

Step 1: Find the number of students with either blond hair or blue eyes

The number of students with neither blond hair nor blue eyes is 8, so: $n(B \cup E) = 45 - 8 = 37$

Step 2: Use the inclusion-exclusion principle

The inclusion-exclusion principle states: $n(B \cup E) = B + E - n(B \cap E)$ Substitute the values: $37 = B + E - 8$ $B + E = 45$

Step 3: Substitute $B = 2E$ into the equation

$2E + E = 45$ $3E = 45$ $E = 15$

Step 4: Check the values

• $B = 2E = 2 \times 15 = 30$
• $n(B \cup E) = 30 + 15 - 8 = 37$
• Number of students with neither: 45 - 37 = 8 (which matches the problem statement).

Conclusion:

The number of students with blue eyes is 15.

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Would you like more details or have any questions? Here are 8 related questions that might interest you:

1. How many students have blond hair but not blue eyes?
2. What is the probability of selecting a student with both blond hair and blue eyes?
3. How would the number change if there were 5 more students with blue eyes?
4. How can we represent this problem using a Venn diagram?
5. What if the ratio of blond hair to blue eyes were different?
6. How many students have only blue eyes?
7. What is the number of students with only blond hair?
8. How would the equation change if 10 students were neither blond nor had blue eyes?

Tip: In problems involving ratios and sets, consider using the principle of inclusion-exclusion to avoid double counting.