This is a Venn diagram problem that can be solved using the inclusion-exclusion principle. Here’s how we analyze and solve it:

Problem Summary:

From a survey of 25 students:

• 15 students took Mathematics (M),
• 12 students took Physics (P),
• 11 students took Chemistry (C),
• 5 students took Mathematics and Chemistry (M ∩ C),
• 9 students took Mathematics and Physics (M ∩ P),
• 4 students took Physics and Chemistry (P ∩ C),
• 3 students took Mathematics, Physics, and Chemistry (M ∩ P ∩ C).

We are tasked to find the number of students in each category described (i) through (viii).

Inclusion-Exclusion Analysis:

Define:

• $|M| = 15, |P| = 12, |C| = 11$
• $|M \cap P| = 9, |M \cap C| = 5, |P \cap C| = 4$
• $|M \cap P \cap C| = 3$
• Total number of students, $|U| = 25$.

Now compute each part:

(i) Only Chemistry:

$|C \setminus (M \cup P)| = |C| - (|M \cap C| + |P \cap C| - |M \cap P \cap C|)$ Substitute values: $|C \setminus (M \cup P)| = 11 - (5 + 4 - 3) = 5$

(ii) Only Mathematics:

$|M \setminus (P \cup C)| = |M| - (|M \cap P| + |M \cap C| - |M \cap P \cap C|)$ Substitute values: $|M \setminus (P \cup C)| = 15 - (9 + 5 - 3) = 4$

(iii) Only Physics:

$|P \setminus (M \cup C)| = |P| - (|M \cap P| + |P \cap C| - |M \cap P \cap C|)$ Substitute values: $|P \setminus (M \cup C)| = 12 - (9 + 4 - 3) = 2$

(iv) Physics and Chemistry but not Mathematics:

$|P \cap C \setminus M| = |P \cap C| - |M \cap P \cap C|$ Substitute values: $|P \cap C \setminus M| = 4 - 3 = 1$

(v) Physics and Mathematics but not Chemistry:

$|P \cap M \setminus C| = |P \cap M| - |M \cap P \cap C|$ Substitute values: $|P \cap M \setminus C| = 9 - 3 = 6$

(vi) Only one of the subjects:

We sum students taking only one subject: $|M \setminus (P \cup C)| + |P \setminus (M \cup C)| + |C \setminus (M \cup P)|$ Substitute values: $4 + 2 + 5 = 11$

(vii) At least one of the three subjects:

This is the total number of students in the union of the sets $M, P, C$. Use the inclusion-exclusion principle: $|M \cup P \cup C| = |M| + |P| + |C| - (|M \cap P| + |M \cap C| + |P \cap C|) + |M \cap P \cap C|$ Substitute values: $|M \cup P \cup C| = 15 + 12 + 11 - (9 + 5 + 4) + 3 = 23$

(viii) None of the subjects:

$|U \setminus (M \cup P \cup C)| = |U| - |M \cup P \cup C|$ Substitute values: $|U \setminus (M \cup P \cup C)| = 25 - 23 = 2$

Final Results:

1. Only Chemistry: 5
2. Only Mathematics: 4
3. Only Physics: 2
4. Physics and Chemistry but not Mathematics: 1
5. Physics and Mathematics but not Chemistry: 6
6. Only one of the subjects: 11
7. At least one of the subjects: 23
8. None of the subjects: 2

Would you like me to provide more details on the calculations or explore similar examples?

Additional Questions:

1. What is the total number of students who take exactly two subjects?
2. How would the calculations change if the total number of students surveyed was different?
3. Can you construct a Venn diagram to represent this problem?
4. How can the inclusion-exclusion principle be applied to four or more sets?
5. What happens if we change the values of any pairwise intersections?

Tip:

When solving such problems, creating a labeled Venn diagram can make understanding intersections and exclusions much clearer.