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In the Mundane Goblet competition, 6 teams compete in a [#permalink]
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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
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Originally posted by zisis on 08 Jul 2010, 14:00.
Last edited by Bunuel on 09 Oct 2012, 04:00, edited 2 times in total.
Edited the question.



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Re: MBAmission: The Quest for 700 [#permalink]
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08 Jul 2010, 14:02
No OA yet: my method:
max points: A B C D E F 15 12 9 6 3 0 = 45POINTS
min points A B C D E F 5 5 5 5 5 5 = 30
4530= 15 points ====> (A)



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Re: MBAmission: The Quest for 700 [#permalink]
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08 Jul 2010, 16:06
zisis 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?
(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 4530=15. Answer: A. Hope it helps.
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Re: MBAmission: The Quest for 700 [#permalink]
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09 Jul 2010, 08:02
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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Re: MBAmission: The Quest for 700 [#permalink]
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06 Aug 2010, 12:14
Its just 3(total teams  1)  0 Coz max pts are if 1 team won all the matches & min pts are 0 In This case 3(61)  0 = 15  0 = 15 HTH
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logical combinatorics [#permalink]
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21 Aug 2011, 09:00
"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 4530=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*150*15=45...or i just missed that there are no any earned scores when the team loses the game?



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Re: logical combinatorics [#permalink]
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21 Aug 2011, 09:11
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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Re: logical combinatorics [#permalink]
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21 Aug 2011, 09:27
good point nammers the points are split between the two teams, and the question says together so it's 2 i.e. a tie
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Re: logical combinatorics [#permalink]
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21 Aug 2011, 18:43
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 4530=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*150*15=45...or i just missed that there are no any earned scores when the team loses the game?



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Re: Mundane Goblet competition [#permalink]
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08 Oct 2012, 09:18
Why we r not taking minumun value 0?? Bunuell..what actually i was doing..i multiplied 3*15=45.. and 15*0 =0...4515=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
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Re: Mundane Goblet competition [#permalink]
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09 Oct 2012, 04:03



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Re: Mundane Goblet competition [#permalink]
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09 Oct 2012, 22:27
sanjoo wrote: Why we r not taking minumun value 0?? Bunuell..what actually i was doing..i multiplied 3*15=45.. and 15*0 =0...4515=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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Re: In the Mundane Goblet competition, 6 teams compete in a [#permalink]
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10 Oct 2012, 03:20
Thanks alot Karishma and Bunuel U both rocked ..
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Re: MBAmission: The Quest for 700 [#permalink]
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14 Jan 2013, 13:51
[quote]
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(n1)/2 which seemed to work okay. Is one method preferable to the other?
Hope my question was clear.



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Re: MBAmission: The Quest for 700 [#permalink]
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14 Jan 2013, 18:40
IanElliott wrote: Quote: 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(n1)/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: \(n!/(nr)!r!\) \(6!/(2!4!)\)



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Re: MBAmission: The Quest for 700 [#permalink]
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14 Jan 2013, 19:42
IanElliott wrote: Quote: 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(n1)/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. mathcombinatorics87345.html
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Re: MBAmission: The Quest for 700 [#permalink]
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14 Jan 2013, 20:44
IanElliott wrote: Quote: 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(n1)/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(n1)/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 For a background on combinations, check this post: http://www.veritasprep.com/blog/2011/11 ... binations/
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Re: MBAmission: The Quest for 700 [#permalink]
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19 Jan 2013, 09:20
IanElliott wrote: Quote: 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(n1)/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) nrepresents 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! /( (nr)! * r !) = (1*2*...*n)/((1*2.....(nr))*(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 = 4530=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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Re: MBAmission: The Quest for 700 [#permalink]
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26 May 2013, 04:16
VeritasPrepKarishma wrote: IanElliott wrote: Quote: 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(n1)/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(n1)/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 For a background on combinations, check this post: http://www.veritasprep.com/blog/2011/11 ... binations/Hi Karishma, Can you please explain why did we add = 5+4+3+2+1 here?




Re: MBAmission: The Quest for 700
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