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In a chess tournament where the participants were to play one game with another, two players fell ill having played 3 games each. If the total number of games played is 84, the number of participants at the beginning was, (game is not held between the two players who fell ill).

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answer is C.

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Detailed Solution

$Let n be the number of players {(n - 2)}_{C}_{2} + 6 = 84 \frac{(n - 2) (n - 3)}{2} = 78 (n - 2) (n - 3) = 156 = 13 (12) \Rightarrow n = 15$

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