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In a tournament with 15 chess players, if each player plays two games

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bmwhype2
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In a tournament with 15 chess players, if each player plays two games [#permalink] New post   21 Nov 2007, 02:40
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In a tournament with 15 chess players, if each player plays two games against every opponent, how many total games will be played?

A. 190
B. 200
C. 210
D. 220
E. 225

M11-29

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C
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Bunuel
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In a tournament with 15 chess players, if each player plays two games [#permalink] New post   23 Feb 2012, 14:12
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In a tournament with 15 chess players, if each player plays two games against every opponent, how many total games will be played?

A. 190
B. 200
C. 210
D. 220
E. 225

The number of unique pairs that can be formed from 15 players is given by \(C^2_{15} = 105\). As each pair competes against each other twice, the total number of games played amounts to \(2 *105 = 210\).

Answer: C.

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Hope it helps.
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Re: In a tournament with 15 chess players, if each player plays two games [#permalink] New post   24 Sep 2009, 20:19
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Though 2*(15C2) is the correct approcah to do this, but for people like me who find Perm, Comb n Prob a nightmare, an easy approach can be used.

The first guy has to play 2 matches with the rest of 14, so he'll play 28 matches.

Similarly, second guy has to play with the rest of 13 as his 2 games with the first guy are already played. So he plays 26 matches.

This continues like this and the total matches are 28+26+24...+2

28+26+...+2 = 2(14+13+...+1) = 2((14*15)/2) = 14*15 = 210.

Hope, it is clear to people like me and the ones who already used the shortkut approach.
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priyankurml
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Re: In a tournament with 15 chess players, if each player plays two games [#permalink] New post   10 Aug 2008, 05:19
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I used simple formula: [n*(n-1)], where n=15; If they play single match, devide the result by 2: [n*(n-1)]/2.

C 210
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Re: In a tournament with 15 chess players, if each player plays two games [#permalink] New post   04 Mar 2010, 03:54
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Hey,
I like the easy approach with 28 x 26 x 24 etc.

However I am wondering why you used the Combinations formula. In my opinion order does matter here and therefore I took the Permutation formula. 15!/(15-2)! = 210, which is also the right solution.

vonshuriz
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Re: In a tournament with 15 chess players, if each player plays two games [#permalink] New post   13 Oct 2010, 23:06
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vonshuriz wrote:
Hey,
I like the easy approach with 28 x 26 x 24 etc.

However I am wondering why you used the Combinations formula. In my opinion order does matter here and therefore I took the Permutation formula. 15!/(15-2)! = 210, which is also the right solution.

vonshuriz


Order does not matter in this situation. For example, if I play a game of chess against you, that is the same as saying you played a game of chess against me. There are 15 players, so pick any 2 to play in a match: 15C2. Now make them play twice: 15C2 * 2.

The Permutation formula worked by luck because everyone played each opponent twice. Had the question asked for everyone to play 3 games, the Permutation formula would not work unless you accounted for double counting: (15P2 / 2!) * 3 = (15C2) * 3
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Re: In a tournament with 15 chess players, if each player plays two games [#permalink] New post   23 Feb 2012, 14:52
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C

15! / (15-2)!
= 210

Initially, I wasnt sure if order matters... Are then any alternate approaches?

Thanks.
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Re: In a tournament with 15 chess players, if each player plays two games [#permalink] New post   23 Feb 2012, 15:16
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rsaraiya wrote:
C

15! / (15-2)!
= 210

Initially, I wasnt sure if order matters... Are then any alternate approaches?

Thanks.


Order has nothing to do with it. The point is that, every person should play twice with each of his opponents and since there are \(C^2_{15}=105\) different pairs of players possible then each pair to play twice 105*2=210 games are to be played.

Check the links in my previous post for similar problems.

Hope it helps.
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Re: In a tournament with 15 chess players, if each player plays two games [#permalink] New post   08 Jan 2014, 15:40
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bmwhype2 wrote:
15 chess players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played?

A. 190
B. 200
C. 210
D. 220
E. 225


15*15 - 15 = 210

Normally one would want to divide by 2, but since they are playing against each other twice, it stands as 210

C

Cheers
J :)
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Re: In a tournament with 15 chess players, if each player plays two games [#permalink] New post   13 Jul 2014, 17:52
jlgdr wrote:

15*15 - 15 = 210

Normally one would want to divide by 2, but since they are playing against each other twice, it stands as 210

C

Cheers
J :)


Please help, How to solve in the above way if the players play 3 matches each?
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Re: In a tournament with 15 chess players, if each player plays two games [#permalink] New post   14 Jul 2014, 03:04
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mankul wrote:
jlgdr wrote:

15*15 - 15 = 210

Normally one would want to divide by 2, but since they are playing against each other twice, it stands as 210

C

Cheers
J :)


Please help, How to solve in the above way if the players play 3 matches each?


In this case the answer would be \(C^2_{15}*3=315\).

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Re: In a tournament with 15 chess players, if each player plays two games [#permalink] New post   05 Jan 2024, 05:32
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