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.