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S96-11

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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?

A. 15
B. 30
C. 45
D. 60
E. 75
[Reveal] Spoiler: OA

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New post 16 Sep 2014, 01:51
Official Solution:

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?

A. 15
B. 30
C. 45
D. 60
E. 75


First, we should determine the number of games played in this competition. We can count them in at least 2 different ways:

(1) Brute force. Name the 6 teams A, B, C, D, E, and F. A plays each of the other teams once, so A plays 5 games. B also plays 5 games, but we've already counted 1 of those games (the game with A), so we have 4 "new" games. C also plays 5 games, but we've already counted 2 of those games (the games with A and with B), so we have 3 "new" games. Continuing, we get \(5 + 4 + 3 + 2 + 1 = 15\) games.

(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 \(\frac{6!}{2!4!} = 15\) games.

Now, to find the maximum and minimum total points earned by all teams in the competition, we should notice that if one team wins and the other team loses, then 3 points total are earned (3 for the win and 0 for the loss). On the other hand, if the game ends in a draw, then only 2 points total are earned (1 by each team). So the maximum total points are earned if every game ends in a win/loss, and the minimum total points are earned if every game ends in a draw.

\(\text{Maximum} = 3 \times 15 = 45\) points.

\(\text{Minimum} = 2 \times 15 = 30\) points.

The difference between the maximum and the minimum is therefore \(45 - 30 = 15\).


Answer: A
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Re S96-11 [#permalink]

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New post 26 Nov 2015, 10:57
I think this the explanation isn't clear enough, please elaborate.
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New post 29 Nov 2015, 20:34
shapla wrote:
I think this the explanation isn't clear enough, please elaborate.


Hi shapla,

let us count the number of games played first..
total 6 teams.. each team plays with each of the other 5 teams..
so total games is ways of choosing 2 teams out of 6 teams=6C2=6!/4!2!=15 games..

now lets see the max and min points in each game..
a) if there is a result as win/loss, total points earned=3(for the winning team ) + 0(for a losing team)=3..
b) if there is a tie = 1+ 1(1 each for both team)=2
therefore the difference is 3-2=1..

there are 15 games, so total difference can be 15*1=15..
A
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New post 30 Nov 2015, 01:16
thanks chetan...
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Re: S96-11 [#permalink]

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New post 04 Oct 2016, 22:29
Can the minimum not be zero? What if there's a team that has lost in every match?

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New post 18 Mar 2017, 14:08
So here is how i approached this problem. I followed the steps below:

a) Create 6x6 matrix
b ) You eliminate the diagonal , now we have : 36-6=30 left
c) Since the teams play just one time with each other we need only 30/2=15 elements
d) 3*15( max points) - 1*2*15(min points and 2 is here because both teams will get points)= 15
So answer : 15

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New post 23 Aug 2017, 19:13
will some one please explain how the minimum score is 30.

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New post 23 Aug 2017, 19:35
Ajstyles wrote:
will some one please explain how the minimum score is 30.

Total number of games played =15 (6C2)

Total minimum score-> every match played was a draw (In every match 1 point was given to both the teams) so total points in that case= 15*2= 30

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Re: S96-11   [#permalink] 23 Aug 2017, 19:35
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