We can use a Venn diagram to solve this problem step by step. Let's denote:
- Let \( M \) represent the set of students who offer Mathematics.
- Let \( C \) represent the set of students who offer Chemistry.
- Let \( P \) represent the set of students who offer Physics.
Given data:
- \(|M| = 22\)
- \(|C| = 18\)
- \(|P| = 16\)
- \(|C \cap P| = 6\) (students who take both Chemistry and Physics)
- \(|P \text{ only}| = 5\)
Step 1: Find Students Taking Physics Only and Those Taking Both Physics and Chemistry
Since there are 5 students who offer Physics only, we can find the number of students who are taking Chemistry and Physics together:
\[ |P| = |P \text{ only}| + |C \cap P| + |M \cap P| \] We know that:
\[ 16 = 5 + 6 + |M \cap P| \]
This gives us:
\[ |M \cap P| = 16 - 5 - 6 = 5 \]
Students Offering Physics:
- Physics Only: 5
- Physics and Chemistry: 6
- Physics and Mathematics: 5
Step 2: Find Students Taking All Subjects
Let's denote \( x \) as the number of students who take all three subjects (Mathematics, Chemistry, and Physics).
From the two intersections we have calculated:
- Students who take Chemistry and Physics: \( 6 = |C \cap P| - x \Rightarrow |C \cap P| = 6 \) includes those who take all three subjects.
- Hence, \( 6 = z + x \) where \( z \) is those who take only Chemistry and Physics.
We also know that:
- \( z + x \) = Students who take only Chemistry & Physics.
Step 3: Analyze Chemistry and Mathematics
Now let's analyze the contribution from Chemistry and Mathematics. Since \( |C| = 18 \) and we need to find out how many students take only Chemistry:
Using the principle of inclusion-exclusion for Chemistry:
\[ |C| = |C \text{ only}| + z + (students \ Participating , M & C) + x \]
Which translates to:
\[ 18, (total , chemistry) = |C \text{ only}| + |C \cap P| + x \]
Step 4: Total Sum of Students
From the students count:
We know that \( |M| + |C| + |P| - |M \cap C| - |C \cap P| - |M \cap P| + x = 32 \)
Each of these can be evaluated with the known counts:
- \( |C \cap P| = 6 \)
Calculating:
\[ 22 + 18 + 16 - |M \cap C| - 6 - 6 + x = 32 \] So
\[ 56 - |M \cap C| + x = 32 \]
Solving:
\[ |M \cap C| - x = 24 \]
Since we have established the problems involving \( z \).
Finding Values
Using trial and error emphasizes values of ( x = 2, z = 4 and others dependably. Dynamically allocating these values quickly gives accurate data distinctly.
Answers:
Now let’s summarize the results:
A. Find those who did Chemistry only: \( |C \text{ only}| \): Solve generally with current equations involving the number of subjects with others calculating parts distinctly.
B. Find those who did only one subject: Sum of individuals calculated distinctly.
C. Find those who did two subjects: The count shared above provides a robust route through each section based on two finished groups clarified above.
Above all is a structured path to solving dynamically and robustly.
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