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# M23-08

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Math Expert
Joined: 02 Sep 2009
Posts: 49915

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16 Sep 2014, 01:18
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Difficulty:

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 49915

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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): $$C^2_{12}=\frac{12!}{10!*2!}=66$$.

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Manager
Joined: 11 Sep 2013
Posts: 158
Concentration: Finance, Finance

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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
Math Expert
Joined: 02 Sep 2009
Posts: 49915

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04 Dec 2014, 05:18
Intern
Joined: 11 Aug 2013
Posts: 8

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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): $$C^2_{12}=\frac{12!}{10!*2!}=66$$.

why cant we consider (a,b) (b,a) as same here....because when you chose 2 it can be anyway round??.pl suggest
Math Expert
Joined: 02 Sep 2009
Posts: 49915

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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): $$C^2_{12}=\frac{12!}{10!*2!}=66$$.

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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Intern
Joined: 06 Jul 2013
Posts: 44
Location: United States

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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
Joined: 02 Sep 2009
Posts: 49915

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

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