Question

In a chess tournament, where the participants were to play one game with another. Two chess players fell ill, having played three games each. If the total number of games played is $84$, the number of participants at the beginning was:

• A. $21$
• B. $16$
• C. $15$
• D. $20$

Answer 1

The correct option is C $15$

Since $2$ participants out of $n$ fell ill, therefore the remaining players are $n-2$

Since the $2$ players had played $6$ games, the remaining games will be $={}^{n-2}{C}_2$
$\Rightarrow {}^{n-2}{C}_2+6=84$
$\Rightarrow \frac{(n-2)(n-3)}{2}+6=84$
$\Rightarrow n^2-5n-150=0$
$\therefore n=15$