Question

In a class tournament when the participants were to play one game with another, two class 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. \(15\)
• B. \(30\)
• C. \(78\)
• D. \(48\)

Hint:

In a tournament where every player plays every other player once, total games are: \[ {}^nC_2=\frac{n(n-1)}{2} \]

Answer 1

The Correct Option is A

Concept:
This is a round-robin tournament problem. If \(n\) players are present and each player has to play exactly one game with every other player, then total number of games is: \[ {}^nC_2=\frac{n(n-1)}{2} \] This is because each game is played between a pair of players.

Step 1: Let the original number of participants be \(n\).
If no one had fallen ill, total number of games would be: \[ {}^nC_2=\frac{n(n-1)}{2} \] But two players fell ill after playing three games each.

Step 2: Check option-based solution.
The options are numerical, so we can verify them directly. Try option (A): \[ n=15 \] Total games if all played: \[ {}^{15}C_2=\frac{15\times14}{2}=105 \]

Step 3: Find how many games were missed.
Two players fell ill after playing only three games each. If there were \(15\) players, each player was supposed to play: \[ 14 \] games. Each ill player played only: \[ 3 \] games. So each ill player missed: \[ 14-3=11 \] games. For two ill players: \[ 2\times 11=22 \] But the game between the two ill players is counted twice in this subtraction if both missed it. So adjust by subtracting one duplicate count: \[ 22-1=21 \] Therefore, missed games: \[ 21 \]

Step 4: Calculate actual games played.
\[ \text{Actual games}=105-21 \] \[ \text{Actual games}=84 \] This matches the question.

Step 5: Final answer.
Therefore, the number of participants at the beginning was: \[ 15 \] Hence, the correct answer is: \[ \boxed{(A)\ 15} \]