In a tournament there are n teams T1,T2,...Tn where n>5. Each team consists of k players, k>3. The following pairs of teams have one player in common:
T1 & T2, T2 & T3, T3 & T4,......... and Tn & T1
No other pairs of teams have any player in common. How many players are participating in the tournament considering all the n teams together?
(1)n(k-1) (2)k(n-1) (3)n(k-2) (4)k(k-2) (5)(n-1)(k-1)
Solution follows here:
There are n teams. Each team has k players. Each team has one player common with two other teams. Each team has “(k-2)” non-common players and 2-common players shared with two teams.
=> Number of overall non-common players = n(k-2) ---(1)
But the number of overall common players is not n*2, as the commonality of each team is with respect to two other teams.
For finding common players, we go in other way. From n teams, n pairs can be formed (T1 & T2, T2 & T3, T3 & T4,......... and Tn & T1). Each of these n-pairs has one player in common.
=> Number of overall common players = n ---(2)
From (1) and (2),
Total number of players = n(k-2)+n = n(k-2+1) = n(k-1)Answer (1)