Problem 52 - Algebra for JEE ADVANCED

Question

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 a. 10 b. 15 c. 12 d. 14

Step-by-Step Solution

Answer

15 participants.

Step 1: Understand the problem

We need to calculate the total number of participants originally in the tournament, where each participant plays a game with every other participant. Two players left after playing 3 games each. A total of 84 games were played.

Step 2: Formulate the equation for games played

For n participants, the number of games played if everyone plays each other once is given by the combination formula: \( \frac{n(n-1)}{2} \). This is the theoretical total if all players remained in the tournament throughout.

Step 3: Adjust for the incapacitated players

Two players left after playing 3 games each, which means a total of 6 games had been played by the two players before leaving. Hence, the remaining games played by others are: \( 84 - 6 = 78 \).

Step 4: Set up the equation

The equation for the games that could be completed without the two incapacitated players: \( \frac{(n-2)(n-3)}{2} = 78 \).

Step 5: Solve the equation for n

\( \frac{(n-2)(n-3)}{2} = 78 \) leads to the quadratic equation \((n-2)(n-3) = 156\). Solving gives us \(n^2 - 5n + 6 = 156\), simplify to \(n^2 - 5n - 150 = 0\).

Step 6: Factor the quadratic equation

Solve \(n^2 - 5n - 150 = 0\) by factoring. The equation factors to \((n - 15)(n + 10) = 0\). Solving this gives \(n = 15\) (ignoring the negative solution).

Step 7: Validate the result

Checking \(n = 15\), the original scenario with all players is \(\frac{15 \times 14}{2} = 105\). After accounting for 6 games by the incapacitated players, 105 - 21 = 84 matches remaining, which aligns with the conditions given.

Key Concepts

Tournament Problems

Tournament problems often entail scenarios where each player, team, or participant competes against every other participant. This type of setup helps us explore various combinatorial concepts including permutations and combinations. Here, the tournament problem is simplified by assuming every player plays exactly once against every other player.

When dealing with such issues, it is crucial to understand the nature of the competition and any constraints, such as players withdrawing or specific numbers of matches. This understanding allows us to set up equations representing the total number of games efficiently.

• If a tournament is arranged as a round-robin, the objective is to compute games played between individuals using combinatorial approaches.
• Understanding the conditions leading to players not completing games helps calculate the net effects on the tournament structure.

Combination Formula

The combination formula is a key part of calculating how many different ways games can be played in a tournament. To find out how many games are played with n participants, each player competes once with each other player. The number of matches can be calculated through the combination formula
<\(\binom{n}{2} = \frac{n(n-1)}{2}\)
which simplifies the process by considering the pairings of players.

• Combinatorics focuses on counting selected items within a set, critical in determining game counts and player interactions during tournaments.
• Utilizing the formula makes it easy to handle the various ways players can face off against each other without the need for lengthy enumeration.

In our tournament example, adjustments were needed to account for players leaving early. Using correct combinations enables efficient tallying of remaining games.

Quadratic Equations

Quadratic equations frequently appear in tournament problems involving combinations. The reason lies in the nature of combinatorial calculations that inherently produce quadratic growth in scenarios involving pairwise games.

The general formula for a quadratic equation is
<\( ax^2 + bx + c = 0 \)
where solving for \(n\) gives the possible values of participants or pairings.

• Factoring the equation or using the quadratic formula yields solutions, which in this context relate to the number of players initially present or remnant after withdrawals.
• The equation's roots supply critical insights, particularly when needing to interpret solutions within realistic constraints (such as the non-acceptance of negative participant numbers).

In our problem, the equation solution \( n = 15 \) represents the starting total before taking into account the players who left, confirming our computations for games left unplayed.