Permutations & Combinations
Counting games/arrangements
Grade 11
Question:
<p>There were two women participating in a chess tournament. Every participant played two games with the other participates. The number of games that the men played among themselves proved to exceed by 66 number of games that the men played with the women. The number of participants is</p>
<p>6</p>
<p>11</p>
<p>13</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Set up equations based on game counts: if there are m men and 2 women, then games among men = m(m-1) and games between men and women = 2m(2), then use the difference condition to solve for m.
<p><strong>Step 1:</strong> Let there be m men and 2 women in the tournament.</p><p><strong>Step 2:</strong> Games among men: Each pair of men plays 2 games. Number of ways to choose 2 men from m men = C(m,2) = m(m-1)/2. Total games among men = 2 × m(m-1)/2 = m(m-1).</p><p><strong>Step 3:</strong> Games between men and women: Each man plays with each woman 2 games. Number of such pairs = m × 2. Total games = 2 × m × 2 = 4m.</p><p><strong>Step 4:</strong> According to the problem: m(m-1) - 4m = 66</p><p><strong>Step 5:</strong> Simplify: m² - m - 4m = 66 ⟹ m² - 5m - 66 = 0</p><p><strong>Step 6:</strong> Factor: (m - 11)(m + 6) = 0 ⟹ m = 11 (since m > 0)</p><p><strong>Step 7:</strong> Total participants = m + 2 = 11 + 2 = 13</p><p>∴ Answer: C (13)</p>
Correct Answer: C