## Friday, 16 December 2011

### Puzzle-29 (CAT-2007)

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:
Solution:
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)

#### 1 comment:

1. 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)

Alternate Method:
Since, (Number of teams is > 5) =>Take, n=6
Since, (Number of players is >3) =>Take, k=4 players per team
From the question,
T1 & T2, T2 & T3, T3 & T4, T4 & T5, T5 & T6, T6 & T1 are the possible pairs.
From the question,
Each pair has 1 player in common.
A] From 1st Pair T1 & T2:
T1 = 4, T2 = 4-1 = 3(Subtract 1, Since 1 player is mutual for T1 & T2)
B] From 2nd Pair T2 & T3:
T2 = 3(Since you have sub T2 in the previous step), T3 = 4-1 = 3(Subtract 1, Since 1 player is mutual for T2 & T3)
C] From 3rd Pair T3 & T4:
T3 = 3, T4 = 4-1 = 3(Subtract 1, Since 1 player is mutual for T3 & T4)
D] From 4th Pair T4 & T5:
T4 = 3, T5 = 4-1 = 3(Subtract 1, Since 1 player is mutual for T4 & T5)
E] From 5th Pair T5 & T6:
T5 = 4, T6 = 4-1 = 3(Subtract 1, Since 1 player is mutual for T5 & T6)
F] From 6th Pair T6 & T1:
T1 = 4, T6 = 3-1 = 2(Subtract 1, Since 1 player is mutual for T1 & T6)
Now, T1 + T2 + T3 + T4 + T5 + T6 = 4+3+3+3+3+2=18, hence 18 is the answer.
Now, Substitute n=6 and k=4 in answer options and see for 18.