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# 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) $n(k - 1)$

Last updated date: 15th Sep 2024
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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.

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.