Permutations & Combinations
Tournament problems
Grade 11
Question:
<p>Ten persons numbered 1, 2, …, 10 play a chess tournament, each player playing against every other player exactly one game. It is known that no game ends in a draw. If \(w_1, w_2, \ldots, w_{10}\) are the number of games won by players 1, 2, 3, …, 10, respectively, and \(l_1, l_2, \ldots, l_{10}\) are the number of games lost by the players 1, 2, …, 10, respectively, then</p>
<p>\(\sum w_i = \sum l_i = 45\)</p>
<p>\(w_i + l_i = 9\)</p>
<p>\(\sum w_i^2 = 81 + \sum l_i^2\)</p>
<p>\(\sum w_i^2 = \sum l_i^2\)</p>
Step-by-Step Solution
Key Concept: In a round-robin tournament with no draws, every game has exactly one winner and one loser. Therefore, the total wins equals total losses equals the total number of games played. This constraint, combined with the fact that each player plays exactly 9 games, creates a system where the sum of all wins must equal the sum of all losses.
<p><strong>Step 1:</strong> Identify the tournament structure. With 10 players, each plays every other player exactly once. Total games = C(10,2) = 45.</p><p><strong>Step 2:</strong> Since no game ends in a draw, each game produces exactly 1 winner and 1 loser. Therefore: Σw_i = 45 and Σl_i = 45.</p><p><strong>Step 3:</strong> For each player i, they play exactly 9 games (against 9 opponents), so: w_i + l_i = 9 for each i = 1, 2, ..., 10.</p><p><strong>Step 4:</strong> Summing across all players: Σ(w_i + l_i) = Σ9 = 90, which gives Σw_i + Σl_i = 90. This confirms: 45 + 45 = 90 ✓</p><p><strong>Key Results (Options A & B):</strong></p><p><strong>A:</strong> Σw_i = Σl_i = 45 ✓ (TRUE - Total wins equal total losses)</p><p><strong>B:</strong> w_i + l_i = 9 for all i ✓ (TRUE - Each player plays exactly 9 games)</p><p>∴ Answer: AB</p>
Correct Answer: AB