A certain organisation has three committees. Only two persons are members of all committees, but every pair of committees have three members in common. What is the least possible number of members on any one committee?

### Concept & Logic Use Set Theory and Venn Diagram principles to distribute common elements among intersecting sets to find the minimum required size of an individual set. ### Step-by-Step Solution * Let the three committees be sets $A, B$, and $C$. * Given: The intersection of all three committees has 2 members. $$|A \cap B \cap C| = 2$$ * Given: Every pair of committees shares 3 members. $$|A \cap B| = 3, |B \cap C| = 3, |C \cap A| = 3$$ * The number of members in exactly two specific committees (for example, A and B only) is calculated by subtracting the central intersection: $$|A \cap B| - |A \cap B \cap C| = 3 - 2 = 1$$ * To find the least possible members in committee $A$, we assume there are $0$ exclusive members in $A$. * Total members in $A = (\text{Only A}) + (\text{Only A & B}) + (\text{Only A & C}) + (\text{All Three})$. * Substitute the values: $$|A| = 0 + 1 + 1 + 2 = 4$$ ### Exam Strategy & Shortcut Visualize a 3-circle Venn diagram. Place the 2 members in the central intersection. Since every overlapping pair shares exactly 3 members, place $3 - 2 = 1$ in each two-circle intersection. A single circle encompasses the center (2) and two adjoining pairs (1 + 1). Minimum size = 4. ### Common Pitfall Double-counting the central intersection. A pair of committees sharing 3 members inherently *includes* the 2 members who are present in all three committees. Failing to subtract those 2 leads to a bloated total. ### Final Answer Therefore, the correct answer is 4.