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OG - 121 - seems right ???

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by AleksandrM » Sun Jun 08, 2008 4:20 am
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A) 15
B) 16
C) 28
D) 56
E) 64

I did not use the approach used in the book. Instead, I did...

8!/2!(8 - 6)! = 8 x 7/2 = 4 x 7 = 28.

Is this a coincedence? :o
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by jasonc » Sun Jun 08, 2008 3:33 pm
either ways work and is correct.
I beat the GMAT! 760 (Q49/V44)
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by chidcguy » Sun Jun 08, 2008 5:25 pm
I am not sure what OG is used. Dont have it handy now

Here is what I did

If A,B,C,D,E,F,G,H are the teams

A can play with 7 teams

B can play with 6 teams (subtract the one thats already counted between A & B)

and so on

7+6+5+4+3+2+1 = 28

In other words the Question is asking how many games are possible with 8 teams and 2 teams playing a game. How many ways can we pick 2 out of 8? 8 C 2, which is what you did
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by Brent@GMATPrepNow » Wed Oct 23, 2019 9:18 am
AleksandrM wrote:There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A) 15
B) 16
C) 28
D) 56
E) 64
There are 8 teams. If we ask each team, "How many teams did you play?" we'll find that each team played 7 teams, which gives us a total of 56 games (since 8 x 7 = 56).

From here we need to recognize that each game has been COUNTED TWICE.
For example, if Team A and Team B play a game, then Team A counts it as a game, and Team B ALSO counts it as a game.

So, to account for the DUPLICATION, we'll divide 56 by 2 to get 28

Answer: C

Cheers,
Brent
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by Scott@TargetTestPrep » Sat Oct 26, 2019 7:46 am
AleksandrM wrote:There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A) 15
B) 16
C) 28
D) 56
E) 64
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).] The situation can therefore be solved by finding the number of combinations of 8 items taken 2 at a time, or 8C2:

8C2 = 8!/[2!(8 - 2)!] = 8!/(2!6!) = (8 x 7)/(2 x 1) = 56/2 = 28

Answer: C

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