Question

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) $n(k - 1)$

Answer 1

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.