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In the Mundane Goblet competition, 6 teams compete in a [#permalink]

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08 Jul 2010, 14:00

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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?

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

There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).

Now, in one game max points (3 points) will be obtained if one team wins and another looses and min points (1+1=2 points) will be obtained if there will be a tie. Hence, maximum points that can be gained by all teams will be 15 games * 3 points=45 and the minimum points that can be gained by all teams will be 15 games * 2 points=30, difference is 45-30=15.

Maybe this is a dumb question but I have to ask. Is \(C^2_6\) the same thing as 6!/(2!*4!)? That’s the only way I could get the 15 by which you multiplied 3 and 2 and then subtracted 30 from 45.
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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?

*15 *30 *45 *60 *75 the decision: 6!/(2!*4!)=15 games Maximum=3*15=45 Minimum=2*15=30, so 45-30=15 the correct answer is A"

Could somebody explain me please why the minimum score i had to calculate as 2*15, instead of 0*15??? I solved it as 3*15-0*15=45...or i just missed that there are no any earned scores when the team loses the game?

When a team wins one team gets 3 points and the other 0 meaning 3 points are earned between the two teams in the match. When the teams draw both teams get a point each meaning 2 points are earned between the teams in the match (hence 2*15).
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the question talks about total points of all teams - meaning they are not just interested in any specific team.

every match is played between two teams - two possible scenarios - one team wins and other loses => total points in that match = 3+0 = 3 its tie - each team gets point => total points in that match = 1+1 = 2

so maximum possible points in any match = 3 and minimum =2

altogether we have 15 games, all 15 can fall in one of those scenarios above.

That's why maximum possible points = 15*3 = 45 minimum possible points = 15*2 = 30

difference is 15.

Galiya wrote:

"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?

*15 *30 *45 *60 *75 the decision: 6!/(2!*4!)=15 games Maximum=3*15=45 Minimum=2*15=30, so 45-30=15 the correct answer is A"

Could somebody explain me please why the minimum score i had to calculate as 2*15, instead of 0*15??? I solved it as 3*15-0*15=45...or i just missed that there are no any earned scores when the team loses the game?

Bunuell..what actually i was doing..i multiplied 3*15=45..

and 15*0 =0...45-15=30..

what i think because if we take 0?? then one team loss is another team win..so it will take same 45 points ...

my concept behind it is rite??

Thanks

To add to what Bunuel said above, we are interested in the total points obtained by all the teams together i.e. the total points earned in all the matches. In any one match, if one team loses and gets 0, the other team wins and gets 3. So, in that match, a total of 3 points are generated. On the other hand, if the two teams draw, they both get a point each and in that game, a total of 2 points are generated. No game can generate less than 2 points. So the minimum points generated by a match are 2 and the maximum are 3. Hence we do 3*15 - 2*15.
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There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).

[end quote]

Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:

Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )

Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.

In my case, I used n(n-1)/2 which seemed to work okay. Is one method preferable to the other?

There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).

[end quote]

Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:

Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )

Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.

In my case, I used n(n-1)/2 which seemed to work okay. Is one method preferable to the other?

Hope my question was clear.

\(C^2_6=15\) is a quick notation to say: a combination (order doesn't matter) of 2 items in a pool of 6 items

In order words: how many possibilities to pick 2 different teams among 6 teams

More generally \(C^r_n\) is a combination of r items in a pool of n items

Which leads to the general formula: \(n!/(n-r)!r!\) \(6!/(2!4!)\)

There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).

[end quote]

Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:

Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )

Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.

In my case, I used n(n-1)/2 which seemed to work okay. Is one method preferable to the other?

Hope my question was clear.

IanElliott, check below link about Combinatorics topic and practice problems. You may find it useful to understand above concepts. math-combinatorics-87345.html _________________

There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).

[end quote]

Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:

Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )

Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.

In my case, I used n(n-1)/2 which seemed to work okay. Is one method preferable to the other?

Hope my question was clear.

There are various ways to get a particular answer. People choose a particular method depending on how they think.

Your thought would have been something like this: There are 6 teams. The first team plays with every other team i.e. with 5 teams and goes away. Now, the second plays with the rest of the 4 teams and goes away. Now the third team plays with the rest of the 3 teams and so on...

So total there are 5 + 4+ 3 + 2 + 1 and you used n(n-1)/2 here.

Another way to think about it is this: There are total 6 teams. Every pair of two teams played one match. Out of 6 teams, we will try to make as many distinct pairs of 2 teams each as possible. For this, we use 6C2 = 6!/2!*4!

There is absolutely nothing wrong with either one of the methods. But you must be able to think in nCr terms too because it comes in very handy in many questions. nCr = choose r items from n

There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).

[end quote]

Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:

Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )

Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.

In my case, I used n(n-1)/2 which seemed to work okay. Is one method preferable to the other?

Hope my question was clear.

Hi ,the method is called permutations and combinations.here we want to find the total number of games played by all teams together.Here we have to choose 2 teams from 6 teams to play game i.e here we are using combinations so the formula to find out the total combination how each team can play with other exactly ones the formula is (n C r) n-represents total teams ,C - combination and r - represents minimum no.of teams required to play a game .In this problem n=6 and r= 2

n C r = n! /( (n-r)! * r !) = (1*2*...*n)/((1*2.....(n-r))*(1*2.....r)) .here 6!/(4! * 2!) = 15.

I hope you understood
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Re: In the Mundane Goblet competition, 6 teams compete in a [#permalink]

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19 Jan 2013, 09:38

total no of games played = 15 max points gained by all teams together = 15* 3 = 45 minimum no.of points gained by all teams together is = 15*2 =30 difference between max and min points = 45-30=15

The procedure to calculate min number of points:

here we require the minimum number of points can be gained by all teams together is 2 points i.e if the game is tied each time will get 1 points making min points each game is 2 points so if all the games are tie ,the minimum points of all games is = 15* 2=30
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......................................................................................... Please give me kudos if my posts help.

There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).

[end quote]

Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:

Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )

Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.

In my case, I used n(n-1)/2 which seemed to work okay. Is one method preferable to the other?

Hope my question was clear.

There are various ways to get a particular answer. People choose a particular method depending on how they think.

Your thought would have been something like this: There are 6 teams. The first team plays with every other team i.e. with 5 teams and goes away. Now, the second plays with the rest of the 4 teams and goes away. Now the third team plays with the rest of the 3 teams and so on...

So total there are 5 + 4+ 3 + 2 + 1 and you used n(n-1)/2 here.

Another way to think about it is this: There are total 6 teams. Every pair of two teams played one match. Out of 6 teams, we will try to make as many distinct pairs of 2 teams each as possible. For this, we use 6C2 = 6!/2!*4!

There is absolutely nothing wrong with either one of the methods. But you must be able to think in nCr terms too because it comes in very handy in many questions. nCr = choose r items from n

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