Question

For general \( n \), how many enemies will each member of \( S \) have?

• A. \( n - 3 \)
• B. \( \frac{1}{2}(n^2 - 3n - 2) \)
• C. \( 2n - 7 \)
• D. \( \frac{1}{2}(n^2 - 5n + 6) \)
• E. \( \frac{1}{2}(n^2 - 7n + 14) \)
• A. \( \frac{1}{2}(n^2 - 5n + 8) \)
• B. \( 2n - 6 \)
• C. \( \frac{1}{2}n(n-3) \)
• D. \( n - 2 \)
• J. \( \frac{1}{2}(n^2 - 7n + 16) \)

Answer 1

The Correct Option is D

Total members in \( S \) = \(\frac{n(n+1)}{2}\). For a given pair \((i,j)\), “friends” are those sharing \( i \) or \( j \). Count of such friends = \((n-1) + (n-1) - 1 = 2n - 3\). Enemies = total members - 1 (itself) - friends = \[ \frac{n(n+1)}{2} - 1 - (2n - 3) = \frac{n^2 + n - 4n + 4}{2} = \frac{n^2 - 3n + 4}{2} - 1 = \frac{n^2 - 5n + 6}{2}. \]

Answer 2

Two members are friends if they share a common index. If both share \( i \) in their pair, then the common friends are all pairs that include \( i \) but not the other member’s distinct index. This yields \( n-2 \) possibilities.