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M23-08

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M23-08  [#permalink]

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New post 16 Sep 2014, 01:18
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A
B
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E

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  15% (low)

Question Stats:

75% (00:45) correct 25% (00:35) wrong based on 144 sessions

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12 teams will play in a tournament. If every team is to play once with each of its opponents, how many games are to be played?

A. 64
B. 66
C. 82
D. 106
E. 132

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Re M23-08  [#permalink]

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New post 16 Sep 2014, 01:18
Official Solution:

12 teams will play in a tournament. If every team is to play once with each of its opponents, how many games are to be played?

A. 64
B. 66
C. 82
D. 106
E. 132

The number of games equals to the total number of different pairs possible out of 12 teams (one game per one pair): \(C^2_{12}=\frac{12!}{10!*2!}=66\).

Answer: B
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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.


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Re: M23-08  [#permalink]

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New post 03 Dec 2014, 10:56
1
We can do this type of problem by always multiplying the last two numbers and then dividing the product by 2

(12*11)/2 = 66
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Re: M23-08  [#permalink]

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New post 04 Dec 2014, 05:18
Raihanuddin wrote:
We can do this type of problem by always multiplying the last two numbers and then dividing the product by 2

(12*11)/2 = 66


Check below:
nine-dogs-are-split-into-3-groups-to-pull-one-of-three-88685.html
in-how-many-different-ways-can-a-group-of-8-people-be-99053.html
6-people-form-groups-of-2-for-a-practical-work-each-group-95344.html
a-group-of-8-friends-want-to-play-doubles-tennis-how-many-55369.html
a-group-of-8-friends-want-to-play-doubles-tennis-how-many-106277.html
in-how-many-different-ways-can-a-group-of-8-people-be-divide-85707.html
how-many-ways-are-there-to-split-a-group-of-6-boys-into-two-105381.html
anthony-and-michael-sit-on-the-six-member-board-of-directors-102027.html
in-how-many-different-ways-can-a-group-of-9-people-be-divide-101722.html

Hope it helps.
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Resources:
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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.


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Re: M23-08  [#permalink]

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New post 16 Sep 2018, 00:48
Bunuel wrote:
Official Solution:

12 teams will play in a tournament. If every team is to play once with each of its opponents, how many games are to be played?

A. 64
B. 66
C. 82
D. 106
E. 132

The number of games equals to the total number of different pairs possible out of 12 teams (one game per one pair): \(C^2_{12}=\frac{12!}{10!*2!}=66\).

Answer: B



why cant we consider (a,b) (b,a) as same here....because when you chose 2 it can be anyway round??.pl suggest
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Re: M23-08  [#permalink]

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New post 16 Sep 2018, 01:14
sdgmat89 wrote:
Bunuel wrote:
Official Solution:

12 teams will play in a tournament. If every team is to play once with each of its opponents, how many games are to be played?

A. 64
B. 66
C. 82
D. 106
E. 132

The number of games equals to the total number of different pairs possible out of 12 teams (one game per one pair): \(C^2_{12}=\frac{12!}{10!*2!}=66\).

Answer: B



why cant we consider (a,b) (b,a) as same here....because when you chose 2 it can be anyway round??.pl suggest


\(C^2_{12}\) gives different groups of 2 out of 12, when order does not matter. So, you'll get only one group with two different teams, say (a,b). Since (a,b) is the same as (b,a) (a playing with b means b is playing with a), we don't need to consider (b,a) here. So, basically the order does not matter.
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Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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.


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M23-08  [#permalink]

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New post 17 Sep 2018, 09:45
Bunuel wrote:
12 teams will play in a tournament. If every team is to play once with each of its opponents, how many games are to be played?

A. 64
B. 66
C. 82
D. 106
E. 132


Hi Bunuel or anyone that could answer this:

Had each team played 3 times instead of once, would the answer be 66*3! Or just 66*3?

+1 for an explanation
Math Expert
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Joined: 02 Sep 2009
Posts: 49915
Re: M23-08  [#permalink]

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New post 18 Sep 2018, 02:28
TippingPoint93 wrote:
Bunuel wrote:
12 teams will play in a tournament. If every team is to play once with each of its opponents, how many games are to be played?

A. 64
B. 66
C. 82
D. 106
E. 132


Hi Bunuel or anyone that could answer this:

Had each team played 3 times instead of once, would the answer be 66*3! Or just 66*3?

+1 for an explanation


Just 66*3: each pair plays thrice.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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?
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Re: M23-08 &nbs [#permalink] 18 Sep 2018, 02:28
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