In a class tournament, all participants were to play different games with one another. Two players fell ill after having played three games each. If the total number of games played in the tournament is equal to 84, the total number of participants in the beginning was equal to

Suppose there are n players in the beginning. The total number of games to be played was equal to ${ }^{\mathrm{n}}{\mathrm{C}}_2$ and each player would have played n-1 games.

Let us assume that A and B fell ill. Now the total number of games of A and B is (n - 1) + (n - 1) - 1 = 2n - 3. But they have played 3 games each. Then their remaining number of games is 2n - 3 - 6 = 2n - 9. Given, 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{ }^{\mathrm{n}}{\mathrm{C}}_2-(2\mathrm{n}-9)=84\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{n}}^2-5\mathrm{n}-150=0\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{n}=15\end{array}$

Alternate solution: 

The number of games excluding A and B is ${ }^{\mathrm{n}-2}{\mathrm{C}}_2$. But before leaving A and B played 3 games each. Then,

${ }^{\mathrm{n}-2}{\mathrm{C}}_2+6=84$

Solving this equation, we get n = 15.