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M23-08

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Bunuel
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M23-08 [#permalink] New post   16 Sep 2014, 01:18
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Question Stats:

80% (00:50) correct 20% (01:07) wrong based on 114 sessions
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12 teams are set to compete in a competition. If every team faces each of its opponents once, how many games will be played in total?

A. 64
B. 66
C. 82
D. 106
E. 132
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B
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Bunuel
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Re M23-08 [#permalink] New post   16 Sep 2014, 01:18
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Official Solution:

12 teams are set to compete in a competition. If every team faces each of its opponents once, how many games will be played in total?

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


The number of games is equivalent to the total number of distinct pairs that can be formed from 12 teams (one game per pair); therefore, the total number of games will be \(C^2_{12} = \frac{12!}{10! * 2!} = 66\).


Answer: B
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Raihanuddin
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Re: M23-08 [#permalink] New post   03 Dec 2014, 10:56
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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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Bunuel
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Re: M23-08 [#permalink] New post   04 Dec 2014, 05:18
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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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Re: M23-08 [#permalink] 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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M23-08 [#permalink] New post   16 Sep 2018, 01:14
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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


\(C^2_{12}\) yields 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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TippingPoint93
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Re: M23-08 [#permalink] 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
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Bunuel
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Re: M23-08 [#permalink] New post   18 Sep 2018, 02:28
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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.
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Re: M23-08 [#permalink] New post   21 May 2020, 04:14
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


Hi Bunuel , Can we also use the logic - First team will pay with 11 other teams , second team with 10 teams , 3rd team with 9 teams and so on .
So Total Matches = 11+10+9+8+7+6+5+4+3+2+1 = 66
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Bunuel
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Re: M23-08 [#permalink] New post   21 May 2020, 04:20
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KARISHMA315 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


Hi Bunuel , Can we also use the logic - First team will pay with 11 other teams , second team with 10 teams , 3rd team with 9 teams and so on .
So Total Matches = 11+10+9+8+7+6+5+4+3+2+1 = 66


Yes, you can do this way too.
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Bunuel
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Re: M23-08 [#permalink] New post   24 Sep 2023, 09:10
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M23-08 [#permalink]
24 Sep 2023, 09:10
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