In a tournament there are n teams T

_{1},T_{2},...T_{n}where n>5. Each team consists of k players, k>3. The following pairs of teams have one player in common:
T

_{1}& T_{2}, T_{2}& T_{3}, T_{3}& T_{4},......... and T_{n}& T_{1}
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 (T

_{1}& T_{2}, T_{2}& T_{3}, T_{3}& T_{4},......... and T_{n}& T_{1}). 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)**
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:

ReplyDeleteT1 & 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.

Hence, answer option is 1.