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

Explanation

Correct option is C)
Let the number of participants be x
If no participant fell ill, then number of games played  \(={ }^{\mathrm{n}} \mathrm{C}_2\)
 
Each player will play n−1 games in the tournament
There n−1 games most include 1 game which should be played by the player with other.

\(\text { No. of games played }={ }^n C_2-[(n-1)+(n-1)-1]+6=84\)

\(\Rightarrow \frac{(\mathrm{n}-1) \mathrm{n}}{2}-[\mathrm{n}-1+\mathrm{n}-1-1]+6=84\)

\(\Rightarrow \frac{(\mathrm{n}-1) \mathrm{n}}{2}-2 \mathrm{n}+3+6=84\)

\(\Rightarrow \frac{(\mathrm{n}-1) \mathrm{n}}{2}-2 \mathrm{n}+9=84\)

\(\Rightarrow \frac{\left(\mathrm{n}^2-\mathrm{n}\right)}{2}-2 \mathrm{n}+9=84\)

\(\begin{aligned} &\Rightarrow \mathrm{n}^2-5 \mathrm{n}-150=0 \\ &\Rightarrow \mathrm{n}^2+10 \mathrm{n}+15 \mathrm{n}-150=0 \\ &\Rightarrow \mathrm{n}(\mathrm{n}+10)-15(\mathrm{n}+10)=0 \\ &\Rightarrow(\mathrm{n}+10)(\mathrm{n}-15)=0 \\ &\mathrm{n}=10[\text { Not possible }] \\ &\mathrm{n}=15 \end{aligned}\)