Question

In a chess tournament, where participants were to play one game with one another, two players fell ill, having played 6 games each without playing among themselves. If the total number of games played was 117, then the number of participants at the beginning was: 

(a) 15 

(b) 16 

(c) 17 

(d) 18

Answer 1

(c) 17

Let the number of participants in the beginning be n. 

Total number of games played by the 2 players who fell ill = 6 + 6 = 12 

Number of remaining players is (n – 2) 

Number of games played by (n – 2) players with each other = ^n–2C₂ 

∴ Total number of games played = ^n–2C₂ + 12 

Given ^n–2C₂ + 12 = 117 

⇒ ^n–2C₂ ⇒ \(\frac{n-2!}{(n-4)!2!}\) = 105

⇒ \(\frac{(n-2)(n-3)}{2}\) = 105 ⇒ n² – 5n – 204 = 0 

⇒ (n – 17) (n + 12) 0 ⇒ n = 17 (Neglecting –ve value)