Find all School-related info fast with the new School-Specific MBA Forum

It is currently 25 Nov 2014, 20:18

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

In the Mundane Goblet competition, 6 teams compete in a

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
Manager
Manager
avatar
Joined: 16 Feb 2010
Posts: 225
Followers: 2

Kudos [?]: 85 [1] , given: 16

In the Mundane Goblet competition, 6 teams compete in a [#permalink] New post 08 Jul 2010, 13:00
1
This post received
KUDOS
3
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

53% (02:02) correct 47% (01:17) wrong based on 178 sessions
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

Last edited by Bunuel on 09 Oct 2012, 03:00, edited 2 times in total.
Edited the question.
1 KUDOS received
Manager
Manager
avatar
Joined: 16 Feb 2010
Posts: 225
Followers: 2

Kudos [?]: 85 [1] , given: 16

Re: MBAmission: The Quest for 700 [#permalink] New post 08 Jul 2010, 13:02
1
This post received
KUDOS
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

45-30= 15 points ====> (A)
Expert Post
9 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 24081
Followers: 3736

Kudos [?]: 31281 [9] , given: 3310

Re: MBAmission: The Quest for 700 [#permalink] New post 08 Jul 2010, 15:06
9
This post received
KUDOS
Expert's post
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 45-30=15.

Answer: A.

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Take a Survey about GMAT Prep - Win Prizes!

Senior Manager
Senior Manager
avatar
Status: Yeah well whatever.
Joined: 18 Sep 2009
Posts: 347
Location: United States
GMAT 1: 660 Q42 V39
GMAT 2: 730 Q48 V42
GPA: 3.49
WE: Analyst (Insurance)
Followers: 5

Kudos [?]: 51 [0], given: 17

Re: MBAmission: The Quest for 700 [#permalink] New post 09 Jul 2010, 07:02
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.
_________________

He that is in me > he that is in the world. - source 1 John 4:4

1 KUDOS received
Current Student
User avatar
Affiliations: Volunteer Operation Smile India, Creative Head of College IEEE branch (2009-10), Chief Editor College Magazine (2009), Finance Head College Magazine (2008)
Joined: 25 Jul 2010
Posts: 471
Location: India
WE2: Entrepreneur (E-commerce - The Laptop Skin Vault)
Concentration: Marketing, Entrepreneurship
GMAT 1: 710 Q49 V38
WE: Marketing (Other)
Followers: 12

Kudos [?]: 87 [1] , given: 24

GMAT ToolKit User
Re: MBAmission: The Quest for 700 [#permalink] New post 06 Aug 2010, 11:14
1
This post received
KUDOS
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(6-1) - 0 = 15 - 0 = 15

HTH
_________________

Kidchaos

http://www.laptopskinvault.com

Follow The Laptop Skin Vault on:
Facebook: http://www.facebook.com/TheLaptopSkinVault
Twitter: http://www.twitter.com/LaptopSkinVault

Consider Kudos if you think the Post is good
Unless someone like you cares a whole awful lot. Nothing is going to change. It's not. - Dr. Seuss

Manager
Manager
User avatar
Joined: 15 Jan 2011
Posts: 107
Followers: 8

Kudos [?]: 46 [0], given: 13

logical combinatorics [#permalink] New post 21 Aug 2011, 08:00
"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
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? :roll:
Manager
Manager
avatar
Joined: 19 Jul 2011
Posts: 100
Concentration: Finance, Economics
Schools: Duke '15
GPA: 3.9
Followers: 0

Kudos [?]: 54 [0], given: 4

Re: logical combinatorics [#permalink] New post 21 Aug 2011, 08:11
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).
_________________

Show Thanks to fellow members with Kudos its shows your appreciation and its free

Manager
Manager
avatar
Status: Meh, I can't take the GMAT before 2017.
Joined: 20 Aug 2011
Posts: 152
Followers: 3

Kudos [?]: 67 [0], given: 0

Re: logical combinatorics [#permalink] New post 21 Aug 2011, 08:27
good point nammers
the points are split between the two teams, and the question says together
so it's 2 i.e. a tie
_________________

Hit kudos if my post helps you.
You may send me a PM if you have any doubts about my solution or GMAT problems in general.

Director
Director
avatar
Joined: 01 Feb 2011
Posts: 767
Followers: 14

Kudos [?]: 59 [0], given: 42

CAT Tests
Re: logical combinatorics [#permalink] New post 21 Aug 2011, 17:43
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
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? :roll:
Senior Manager
Senior Manager
avatar
Joined: 06 Aug 2011
Posts: 409
Followers: 2

Kudos [?]: 68 [0], given: 82

Re: Mundane Goblet competition [#permalink] New post 08 Oct 2012, 08:18
Why we r not taking minumun value 0??

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 :)
_________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 24081
Followers: 3736

Kudos [?]: 31281 [0], given: 3310

Re: Mundane Goblet competition [#permalink] New post 09 Oct 2012, 03:03
Expert's post
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...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 :)


How can we have zero points in one game? The minimum points in one game is 1+1=2 points, when there is a tie between the teams.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Take a Survey about GMAT Prep - Win Prizes!

Expert Post
2 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4948
Location: Pune, India
Followers: 1184

Kudos [?]: 5639 [2] , given: 167

Re: Mundane Goblet competition [#permalink] New post 09 Oct 2012, 21:27
2
This post received
KUDOS
Expert's post
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...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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Senior Manager
Senior Manager
avatar
Joined: 06 Aug 2011
Posts: 409
Followers: 2

Kudos [?]: 68 [0], given: 82

Re: In the Mundane Goblet competition, 6 teams compete in a [#permalink] New post 10 Oct 2012, 02:20
Thanks alot Karishma and Bunuel :) U both rocked :)..
_________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Intern
Intern
avatar
Joined: 04 Jan 2013
Posts: 3
Followers: 0

Kudos [?]: 1 [0], given: 11

Re: MBAmission: The Quest for 700 [#permalink] New post 14 Jan 2013, 12:51
[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(n-1)/2 which seemed to work okay. Is one method preferable to the other?


Hope my question was clear.
1 KUDOS received
Intern
Intern
avatar
Joined: 30 Apr 2011
Posts: 10
Followers: 0

Kudos [?]: 1 [1] , given: 0

Re: MBAmission: The Quest for 700 [#permalink] New post 14 Jan 2013, 17:40
1
This post received
KUDOS
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(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! :arrow: 6!/(2!4!)
1 KUDOS received
Current Student
User avatar
Joined: 27 Jun 2012
Posts: 418
Concentration: Strategy, Finance
Followers: 50

Kudos [?]: 388 [1] , given: 182

Re: MBAmission: The Quest for 700 [#permalink] New post 14 Jan 2013, 18:42
1
This post received
KUDOS
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(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
_________________

Thanks,
PraPon

Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7
Reading Comprehension notes: Click here
VOTE: vote-best-gmat-practice-tests-excluding-gmatprep-144859.html
PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here

Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4948
Location: Pune, India
Followers: 1184

Kudos [?]: 5639 [1] , given: 167

Re: MBAmission: The Quest for 700 [#permalink] New post 14 Jan 2013, 19:44
1
This post received
KUDOS
Expert's post
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(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

For a background on combinations, check this post:
http://www.veritasprep.com/blog/2011/11 ... binations/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Intern
Intern
User avatar
Joined: 16 Nov 2012
Posts: 44
Location: United States
Concentration: Operations, Social Entrepreneurship
Schools: ISB '15, NUS '16
GMAT Date: 08-27-2013
GPA: 3.46
WE: Project Management (Other)
Followers: 0

Kudos [?]: 13 [0], given: 54

Reviews Badge
Re: MBAmission: The Quest for 700 [#permalink] New post 19 Jan 2013, 08:20
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(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
_________________

.........................................................................................
Please give me kudos if my posts help.

Intern
Intern
User avatar
Joined: 16 Nov 2012
Posts: 44
Location: United States
Concentration: Operations, Social Entrepreneurship
Schools: ISB '15, NUS '16
GMAT Date: 08-27-2013
GPA: 3.46
WE: Project Management (Other)
Followers: 0

Kudos [?]: 13 [0], given: 54

Reviews Badge
Re: In the Mundane Goblet competition, 6 teams compete in a [#permalink] New post 19 Jan 2013, 08:38
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
_________________

.........................................................................................
Please give me kudos if my posts help.

Manager
Manager
avatar
Joined: 14 Nov 2011
Posts: 145
Location: United States
Concentration: General Management, Entrepreneurship
Schools: Stanford '15
GPA: 3.61
WE: Consulting (Manufacturing)
Followers: 0

Kudos [?]: 20 [0], given: 97

GMAT ToolKit User
Re: MBAmission: The Quest for 700 [#permalink] New post 26 May 2013, 03:16
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(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

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   [#permalink] 26 May 2013, 03:16
    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic In a kickball competition of 9 teams, how many possible ma amgelcer 3 19 Oct 2013, 05:46
10 Experts publish their posts in the topic In a business school case competition, the top three teams alchemist009 7 05 Jul 2012, 22:02
4 Of the teams competing in the world archery championships, twenty perc virtualanimosity 7 20 Aug 2009, 13:48
3 7 teams compete in a track competition. If there are 20 bmwhype2 3 19 Feb 2008, 09:45
3 teams which each team has 3 people attend the competition, gamjatang 3 11 Dec 2005, 05:34
Display posts from previous: Sort by

In the Mundane Goblet competition, 6 teams compete in a

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page    1   2    Next  [ 21 posts ] 



GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.