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Challenge Question 6/28/10 - Mundane Goblet

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marc.gagnon
 Post subject: Challenge Question 6/28/10 - Mundane Goblet
 Post Posted: Wed May 25, 2011 10:59 pm 
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Students


Posts: 17
Can you help me understand what the 2 and the 4 represent in the cobinatorics denominator.


In the Mundane Goblet competition, 6 teams compete in a “round robin” format: that is, each team plays every other team exactly once. A team gets 3 points for a win, 1 point for a tie (a draw), and 0 points for a loss. What is the difference between the maximum total points and the minimum total points that can be gained by all teams (added together) in the Mundane Goblet competition

Explanation provided:

First, we should determine the number of games played in this competition. We can count them in at least 2 different ways:
Method (1) Brute Force
Method (2) Combinatorics. We have a pool of 6 teams, and we want to count how many different pairs of teams (to play a game) we can select, without caring about order. Using either the anagram method or the formula for combinations, we get 6!/(2!4!) = 15 games
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george.kourdin
 Post subject: Re: Challenge Question 6/28/10 - Mundane Goblet
 Post Posted: Thu May 26, 2011 9:37 am 
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Course Students


Posts: 98
combiantions formula is nCr or n!/ r!(n-r)!

we want to figure out how many possible pair arrangements (2 teams playing each other) we can get from a pool of 6 teams. we don't care exactly how these teams are going to be arranged, we are just trying to figure out some n number of arrangements from a population of 6 teams.

n=6 = our population
r = 2 = our pair or group of 2
n-r = 4
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marc.gagnon
 Post subject: Re: Challenge Question 6/28/10 - Mundane Goblet
 Post Posted: Fri May 27, 2011 8:54 am 
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Students


Posts: 17
Thanks!
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jnelson0612
 Post subject: Re: Challenge Question 6/28/10 - Mundane Goblet
 Post Posted: Sat May 28, 2011 3:34 pm 
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ManhattanGMAT Staff


Posts: 2390
Great! :-)

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Jamie Nelson
ManhattanGMAT Instructor
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