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In a tournament, each participant should play one game with other .Two participants fell ill after playing 3 games.each. If total games played is 84, the no. of participants in the tournament is ?

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Let the no. of participants be $n$.
If no participant fell ill, then the no. of games played = $^nC_2$
Each player will play $n-1$ games in the tournament.
These n-1 games include 1 game which both the player played with each other.
This one game is counted with both the players.
The two participants played 3 games each and fell ill.
$\therefore$ The no. of games played = $^nC_2-[(n-1)+(n-1)-1]+6=84$ (given)
$\Rightarrow\:\large\frac{n(n-1)}{2}$$-2n+9=84$
$\Rightarrow\:n^2-5n-150=0$
$\Rightarrow\:n=15\:or\:-10 \:(not\: possible)$
$\Rightarrow\:n=15$
answered Aug 27, 2013 by rvidyagovindarajan_1
edited Aug 6, 2014 by sharmaaparna1
 

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