Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

In a tournament, there are \[n\] teams, \[{T_1},{T_2},...,{T_n}\] with \[n > 5\]. Each team consists of k players, \[k > 3\]. The following pairs of teams have one player common \[{T_1}\] and \[{T_2}\], \[{T_2}\] and \[{T_3}\],…, \[{T_{n - 1}}\] and \[{T_n}\], \[{T_n}\] and \[{T_1}\]. No other pair of teams has many players in common. How many players are participating in the tournament, considering all the \[n\] teams together?
A) \[k(n - 1)\]
B) \[n(k - 3)\]
C) \[n(k - 2)\]
D) \[n(k - 1)\]

seo-qna
Last updated date: 13th Jun 2024
Total views: 403.2k
Views today: 6.03k
Answer
VerifiedVerified
403.2k+ views
Hint: At first, we will find the total number of players. Since, we have \[n\] number of common players, so we will subtract the number of common players to get exact numbers of players. Finally, we will get the exact number of players.

Complete step-by-step answer:
It is given that; In a tournament, there are \[n\] teams, \[{T_1},{T_2},...,{T_n}\]with \[n > 5\]. Each team consists of k players, \[k > 3\]. The following pairs of teams have one player common \[{T_1}\] and \[{T_2}\], \[{T_2}\] and \[{T_3}\],…, \[{T_{n - 1}}\] and \[{T_n}\], \[{T_n}\] and \[{T_1}\]. Moreover, no other pair of teams has many players in common.
We have to find the number of players participating in the tournament, considering all the \[n\] teams together.
Since, the number of teams is \[n\] and number of players \[k\].
So, the total number of players are \[ = n \times k\].
Further, the following pairs of teams have one player common \[{T_1}\] and \[{T_2}\], \[{T_2}\] and \[{T_3}\],…, \[{T_{n - 1}}\] and \[{T_n}\], \[{T_n}\] and \[{T_1}\]. So, the number of common players is \[n\] since, there are \[n\] numbers of teams.
Since, the total number of players are \[ = n \times k\] and we have \[n\] number of common players, so we will subtract the number of common players to get exact numbers of players.
So, the number of exact numbers of players is \[ = n \times k - n = n(k - 1)\]
Hence, the number of exact numbers of players is \[n(k - 1)\].

The correct option is D) \[n(k - 1)\].

Note: It is given, the number of teams should be greater than 5.
Let us consider, the number of teams is 7.
So, the number of common players in 7 teams is
1 player …………. \[{T_1}\] and \[{T_2}\]
1 player …………. \[{T_2}\] and \[{T_3}\]
1 player …………. \[{T_3}\] and \[{T_4}\]
1 player …………. \[{T_4}\] and \[{T_5}\]
1 player …………. \[{T_5}\] and \[{T_6}\]
1 player …………. \[{T_6}\] and \[{T_7}\]
1 player …………. \[{T_7}\] and \[{T_1}\]
This is an example that, if we choose 7 teams, we will get 7 common players.