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If 8 teams participate in Indian Premier League and every te

by GMATinsight » Sun May 07, 2017 8:09 pm
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

Source: www.GMATinsight.com

Answer: Option C
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by Brent@GMATPrepNow » Mon May 08, 2017 4:50 am
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
Let the 8 teams be A, B, C, D, E, F, G, and H

One approach is to ask each team how many games it will play.
For example, team A plays each of the 7 other teams (B, C, D, E, F, G, and H) two times each.
So, team A plays 14 games

In fact, each team plays 14.
So, the total number of games = (8)(14) = 112

IMPORTANT: Each game has been counted TWICE For example, when Team A says it will play 14 games, this includes the games played against Team B. Likewise, when Team B says it will play 14 games, this includes the games to be played against Team A.

To account for the duplication, we'll divide 112 by 2 to get 56

Answer: C

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by Matt@VeritasPrep » Thu May 11, 2017 8:51 pm
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.
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by Jeff@TargetTestPrep » Fri May 12, 2017 12:22 pm
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

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