<p>In a class tournament, all participants were to play different games with one another. Two players fell ill after having played three games each. If the total number of games played in the tournament is equal to 84, the total number of participants in the beginning was equal to</p>
Step-by-Step Solution
Key Concept: Set up an equation where the total games = games among healthy players + games played by the two ill players before leaving. The two ill players played 3 games each (some possibly with each other), which must be subtracted from the complete round-robin total.
<p><strong>Step 1:</strong> Let n = total number of participants initially.</p><p><strong>Step 2:</strong> In a complete round-robin tournament, total games = C(n,2) = n(n-1)/2.</p><p><strong>Step 3:</strong> Two players fell ill after playing 3 games each. They played a total of 6 games, but one of these games might be between themselves (counted once). So we subtract 6 games from the complete tournament.</p><p><strong>Step 4:</strong> After two players leave, (n-2) players continue and play a complete round-robin among themselves = C(n-2,2) = (n-2)(n-3)/2.</p><p><strong>Step 5:</strong> The 2 ill players together contributed: 3 + 3 = 6 games before leaving.</p><p><strong>Step 6:</strong> Total games played = C(n-2,2) + 6 = 84</p><p><strong>Step 7:</strong> (n-2)(n-3)/2 + 6 = 84</p><p><strong>Step 8:</strong> (n-2)(n-3)/2 = 78</p><p><strong>Step 9:</strong> (n-2)(n-3) = 156</p><p><strong>Step 10:</strong> Let m = n-2, then m(m-1) = 156, so m² - m - 156 = 0</p><p><strong>Step 11:</strong> m² - m - 156 = (m-13)(m+12) = 0, thus m = 13</p><p><strong>Step 12:</strong> n = m + 2 = 15</p><p><strong>Step 13:</strong> Verification: C(13,2) + 6 = 78 + 6 = 84 ✓</p><p>∴ Answer: D (15 participants)</p>
Correct Answer: D