Soccer time

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by Brent@GMATPrepNow » Sun Nov 09, 2014 1:06 pm
There are 16 teams in a soccer league, and each team plays each of the others once. Given that each game is played by two teams, how many total games will be played?

A) 256
B) 230
C) 196
D) 169
E) 120
Another approach (other than 16C2) is this:

If we ask each team, "How many teams did you play?" we'll find that each team played 15 teams, which gives us a total of 240 games (since 16 x 15 = 240).

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 counts it as a game.

So, to account for the duplication, we'll divide 240 by 2 to get [spoiler]120 (E)[/spoiler]
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by Matt@VeritasPrep » Mon Nov 10, 2014 10:59 am
Another nice way to see this quickly on test day is using a small set.

For instance, suppose that A, B, and C are playing each other exactly once.

With three teams, it's easy to list all the matchups: A vs B, A vs C, and B vs C.

If we do 3 * 2, however, we get SIX matchups. That means we must be counting each matchup TWICE that way, so we divide by 2. Done!

From here, we can at least guess that any number of teams would work the same way: (n*(n-1))/2.

This sort of pattern hunting works extremely well on the GMAT, I think.

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by nikhilgmat31 » Mon Jul 13, 2015 1:33 am
E is the answer.

first team start 15 games.
second team start 14 games.
....so on

15+14+13+12+ ....+1 = 120

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by KAUSTAV89 » Sun Oct 16, 2016 3:21 am
Imagine a polygon of 16 sides. Each vertex stands for one team.
Draw all the diagonals.
Each side / each diagonal stands for a match between two teams (two vertices at the two end of the line)

Hence total no of matches = total no of diagonals + no of sides

= n(n-3)/2 + n

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by [email protected] » Sun Oct 16, 2016 10:05 am
Hi All,

This question can be answered in a couple of different ways. If you know the Combination Formula, then you can use that...

N!/[K!(N-K)!] where N is the total number of Teams and K is the Subgroup.

In this prompt, N = 16 and K = 2...

16!/[2!(14!)] =

(16)(15)/(2)(1) =

120 different games played

You can also use 'brute force' and a bit of logic to answer the question....

Let's call the teams...ABCDE FGHIJ KLMNO P

Team A plays each of the other 15 teams, so that's 15 games.
Team B already played Team A, so it plays 14 OTHER games.
Team C already played Teams A and B, so it plays 13 OTHER games.
Team D already played Teams A, B and C, so it plays 12 OTHER games.
Etc.

The sum of all of these games is...
15+14+13.....+3+2+1 = 120

Final Answer: E

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by deepak4mba » Fri Apr 06, 2018 12:54 am