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) $$n(k - 1)$$

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) $$n(k - 1)$$

Vedantu Content Team · Accepted Answer

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 is1 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.