To solve the problem using sets and a Venn diagram, let's define the sets and then calculate the number of distinct candidates.

### Definitions of Sets:
- Let \( E \) represent the set of candidates running for the EVP position.
- Let \( A \) represent the set of candidates running for the VP-Academics position.
- Let \( C \) represent the set of candidates running for the VP-Activities position.

### Given Information in Set Terms:
1. \(|E| = 5\) (Total candidates for EVP)
2. \(|A| = 6\) (Total candidates for VP-Academics)
3. \(|C| = 5\) (Total candidates for VP-Activities)
4. \( |E \cap A \cap C| = 2\) (2 candidates are in all three sets)
5. \( |A \cap C| - |E \cap A \cap C| = 3 \) (3 candidates are in both VP-Academics and VP-Activities but not EVP, so \( |A \cap C| = 3 + 2 = 5\))
6. \( |E \setminus (A \cup C)| = 2 \) (2 candidates are running for EVP alone)

### Steps to Solve:
1. **Set Intersection Calculations**:
   - \( |E \cap A \cap C| = 2 \)
   - \( |A \cap C| = 5 \), therefore \( |A \cap C \setminus E| = 5 - 2 = 3 \)
   - \( |E \setminus (A \cup C)| = 2 \)

2. **Calculate Each Segment in the Venn Diagram**:
   - Candidates running only for EVP: \( |E \setminus (A \cup C)| = 2 \)
   - Candidates running for all three positions: \( |E \cap A \cap C| = 2 \)
   - Candidates running for VP-Academics and VP-Activities but not EVP: \( |A \cap C \setminus E| = 3 \)
   - Candidates running only for VP-Academics (not in \(C\) or \(E\)): \( |A \setminus (E \cup C)| \). From \( |A| = 6 \), subtract those who are also in \(C\) or \(E\): \( 6 - (3 + 2 + 1) = 1 \), so \( |A \setminus (E \cup C)| = 1 \).
   - Candidates running for EVP and one other position but not the third: There’s 1 candidate left, which is running for \( EVP \) and either \( A \) or \( C \) (not specified).

3. **Sum the Unique Individuals**:
   - \( |E \setminus (A \cup C)| = 2 \) (EVP only)
   - \( |E \cap A \cap C| = 2 \) (All three positions)
   - \( |A \cap C \setminus E| = 3 \) (VP-Academics and VP-Activities, not EVP)
   - \( |A \setminus (E \cup C)| = 1 \) (VP-Academics only)
   - The remaining candidate falls into a single overlap not covered (either \( |E \cap A| \) or \( |E \cap C| \)).

Therefore, the total number of distinct individuals running for officerships is:

\[
2 \text{ (EVP only)} + 2 \text{ (all three positions)} + 3 \text{ (VP-Academics and VP-Activities only)} + 1 \text{ (VP-Academics only)} + 1 \text{ (the remaining candidate)} = 9
\]

### Visual Representation:
- **Set \(E\)** contains 2 in all overlaps and 2 only in EVP.
- **Set \(A\)** contains 2 in all overlaps, 3 in overlap with \(C\) only, and 1 exclusively in \(A\).
- **Set \(C\)** contains similar overlaps, with remaining candidate potentially in \(E\cap A\) or \(E\cap C\).

### Conclusion:
Using set theory and Venn diagrams, the solution confirms that **9 distinct candidates** are running for officerships.

Would you like more details or have any questions?

### Further Questions:
1. How does a Venn diagram simplify understanding overlapping sets?
2. Can you create a more complex problem using 4 positions and solve similarly?
3. How would this change if one candidate withdrew?
4. How does this method compare to the inclusion-exclusion principle directly?
5. How does this relate to probability problems involving events and outcomes?

**Tip:** Practicing with Venn diagrams can help you better visualize and solve complex set problems.

Explore how to solve a complex set theory problem using sets and Venn diagrams. This problem involves calculating the number of distinct candidates running for various positions based on intersection and set differences. Learn step-by-step with detailed explanations.

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