In a chess tournament, where participants were to play one game with o

1.	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» (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)

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»

(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)