Question

In a class tournament where the participants were to play one game with another, two class 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 :

• A. 15
• B. 16
• C. 20
• D. 21

Answer 1

Answer: (A)

Suppose there were $n$ participants in the beginning. Then the number of games played by $(n-2)$ players $={}^{n-2}{C}_2$
$\therefore$ by the given condition ${}^{n-2}{C}_2+6=84$
($\because$ two players played $3$ games each)
$\therefore {}^{n-2}{C}_2=78$
$\Rightarrow \left(n-2\right)\left(n-3\right)=156$
$\Rightarrow n^2-5n+6=156$
i.e., $n^2-5n-150=0$
$\Rightarrow \left(n-15\right)\left(n+10\right)=0$
$\Rightarrow n=15$
[$\because n$ cannot be $-ve$]