Let’s solve the problem step-by-step:

Part (a) Proof of the formula for $\text{Card}(A \cup B \cup C)$:

We need to show that: $\text{Card}(A \cup B \cup C) = \text{Card}(A) + \text{Card}(B) + \text{Card}(C) - \text{Card}(A \cap B) - \text{Card}(A \cap C) - \text{Card}(B \cap C) + \text{Card}(A \cap B \cap C).$

Explanation:

1. Step 1: Start with the principle of inclusion-exclusion. For two sets, $A$ and $B$, we have: $\text{Card}(A \cup B) = \text{Card}(A) + \text{Card}(B) - \text{Card}(A \cap B).$
2. Step 2: Extend this to three sets. Adding $C$, the union $A \cup B \cup C$ adds $\text{Card}(C)$, but subtracts the pairwise intersections already counted in $\text{Card}(A)$, $\text{Card}(B)$, and $\text{Card}(C)$. Hence: $\text{Card}(A \cup B \cup C) = \text{Card}(A) + \text{Card}(B) + \text{Card}(C) - \text{Card}(A \cap B) - \text{Card}(A \cap C) - \text{Card}(B \cap C).$
3. Step 3: Add back the triple intersection $\text{Card}(A \cap B \cap C)$, which was subtracted too many times. The final result is: $\text{Card}(A \cup B \cup C) = \text{Card}(A) + \text{Card}(B) + \text{Card}(C) - \text{Card}(A \cap B) - \text{Card}(A \cap C) - \text{Card}(B \cap C) + \text{Card}(A \cap B \cap C).$

Part (b) How many students took none of the three specialties?

We are given:

• Total students: $\text{Card}(E) = 36$,
• $\text{Card}(M) = 13$, $\text{Card}(SP) = 10$, $\text{Card}(SVT) = 7$,
• Pairwise intersections:
  • $\text{Card}(M \cap SP) = 6$,
  • $\text{Card}(M \cap SVT) = 2$,
  • $\text{Card}(SP \cap SVT) = 3$,
• Triple intersection: $\text{Card}(M \cap SP \cap SVT) = 2$.

We calculate $\text{Card}(M \cup SP \cup SVT)$ using the inclusion-exclusion principle: $\text{Card}(M \cup SP \cup SVT) = \text{Card}(M) + \text{Card}(SP) + \text{Card}(SVT) - \text{Card}(M \cap SP) - \text{Card}(M \cap SVT) - \text{Card}(SP \cap SVT) + \text{Card}(M \cap SP \cap SVT).$ Substitute the values: $\text{Card}(M \cup SP \cup SVT) = 13 + 10 + 7 - 6 - 2 - 3 + 2 = 21.$

Finally, the number of students who took none of these specialties is: $\text{Card}(\text{None}) = \text{Card}(E) - \text{Card}(M \cup SP \cup SVT).$ Substitute: $\text{Card}(\text{None}) = 36 - 21 = 15.$

Final Answer:

15 students did not take any of the three specialties.

Would you like me to provide detailed Venn diagram explanations or examples? Here are related questions:

1. What is the general formula for the union of $n$ sets using inclusion-exclusion?
2. How do we calculate intersections involving more than three sets?
3. Can you show an example where all pairwise intersections are empty?
4. How does this principle apply in probability theory?
5. How would the result change if the total number of students changed?

Tip: When solving inclusion-exclusion problems, carefully account for every overlap to avoid double-counting.