Question

In a tournament, there are \( n \) teams \( T_1, T_2, \ldots, T_n \), with \( n > 5 \). Each team consists of \( k \) players, \( k > 3 \). The following pairs of teams have one player in common: \( T_1 \) & \( T_2 \), \( T_2 \) & \( T_3 \), \( \ldots \), \( T_{n-1} \) & \( T_n \), \( T_n \) & \( T_1 \). No other pair of teams has any player in common. How many players are participating in the tournament, considering all the \( n \) teams together?  

• A. \( n(k-1) \)
• B. \( k(n-1) \)
• C. \( n(k-2) \)
• D. \( k(n-2) \)
• E. \( (n-1)(k-1) \)

Hint:

In counting participants with overlaps, use inclusion-exclusion: subtract duplicates caused by shared players.

Answer 1

The Correct Option is A

To solve the problem, we need to determine the number of unique players participating in the tournament. We have \( n \) teams, each with \( k \) players, and pairs of teams share exactly one player according to the given pattern.

 Step-by-step Explanation: 

1. Each team \( T_i \) has \( k \) players.
2. Each pair of consecutive teams shares exactly one player. For example, teams \( T_1 \) and \( T_2 \) share one player, while teams \( T_2 \) and \( T_3 \) also share one (not necessarily the same) player.
3. This pattern continues such that team \( T_n \) and team \( T_1 \) also share exactly one player.
4. Visualize the teams in a circular arrangement. The sequence of shared players forms a continuous loop around the circle of teams.
5. Consider the overall distribution: if each team of \( n \) teams shares players with its neighbor, then each team initially contributes all \( k \) players, but each team is overlapping with its neighboring teams by one player per connection.
6. To find unique players, subtract the overlapping players: since there are \( n \) pairs of teams sharing a player, \( n \) players are overlapping and subtracted.
7. Hence, total unique players across all teams can be calculated by subtracting the \( n \) overlaps from the total players initially counted: \[ (n \times k) - n = n(k-1) \]

Therefore, the total number of unique players in the tournament is \( n(k-1) \).