We've got eight teams and we want to choose all possible groups of two teams. This is a combination, 8 choose 2, so we turn to our combinations formula:

x choose y = x! / (y! * (x - y)!)

8 choose 2 = 8! / 2!6!

= 8*7/2

= 28

Each match will be played twice, so we double this: 28 * 2, or 56.

GMATinsight wrote:If 8 teams participate in Indian Premier League and every team is scheduled to play exactly two matches with every other team in the first round then find total number of matches to be played in the first round???

A) 28
B) 49
C) 56
D) 64
E) 112

We are given that there are 8 teams in a league and that each game is played by 2 teams. Note that each team does not play itself and the order of pairing each team with its opponent doesn't matter. [For example, the pairing of (Team A vs. Team B) is identical to the pairing of (Team B vs. Team A).]

Thus, let's first determine the number of games played if each team played against another team just once:

8C2 = 8! / [2! x (8-2)!]

(8 x 7 x 6!) / (2! x 6!)

(8 x 7)/2!

(8 x 7) / 2

4 x 7 = 28

Since each team plays two matches with every other team, the total games played is 28 x 2 = 56.

Alternate Solution:

The first team will play 7 x 2 = 14 games since there are 7 teams besides the first team and 2 games are played with each.

The second team, not counting the games already played with the first team, will play 6 x 2 = 12 games since there are 6 teams besides the first and the second teams.

The third team, not counting the games already played with the first and second teams, will play 5 x 2 = 10 games since there are 5 teams besides the first, second, and third teams.

Following the pattern, we will find that the fourth, fifth, sixth, and seventh teams will play 4 x 2 = 8 games, 3 x 2 = 6 games, 2 x 2 = 4 games, and 1 x 2 = 2 games, respectively.

In total, 14 + 12 + 10 + 8 + 6 + 4 + 2 = 56 games will be played.

Answer: C