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# 15 chess players take part in a tournament. Every player

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SVP
Joined: 04 May 2006
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15 chess players take part in a tournament. Every player [#permalink]  12 May 2008, 23:59
15 chess players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played?

190
200
210
220
225

The OE is difficult for me to understand!
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CEO
Joined: 17 May 2007
Posts: 2994
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Kudos [?]: 468 [0], given: 210

Re: M11-29 [#permalink]  13 May 2008, 00:09
C

Quite easy.

Whenever there is a problem such as this where there are n participants every participant plays everyone else , you can calculate the total number of games by the the series 1 + 2 + 3 + 4 + .. n-1

So in this case this series adds to 14 * 7.5 = 105 games. If they play each other twice, its 105 * 2 = 210 games.

sondenso 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?

190
200
210
220
225

The OE is difficult for me to understand!
VP
Joined: 10 Jun 2007
Posts: 1465
Followers: 6

Kudos [?]: 133 [0], given: 0

Re: M11-29 [#permalink]  13 May 2008, 05:38
sondenso 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?

190
200
210
220
225

The OE is difficult for me to understand!

C

Imagine you have 4 people, ABCD, and each player play with each other once.
In each arrangements, you have
AB, AC, AD => 3 total
BC, BD => 2 total
CD => 1 total
Total = 3 + 2 + 1
Do the same for 15 people...and you will have
14+13+12+...+1 = 105
Twice that, you get 105*2 = 210
Current Student
Joined: 28 Dec 2004
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Re: M11-29 [#permalink]  13 May 2008, 10:28
yeah..i too get 210..

general formula for these type of problems is sum of 1 to (n-1)*Number of times each player plays a game..
Manager
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Location: São Paulo
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Re: M11-29 [#permalink]  13 May 2008, 15:04
I did with a different formula:
15!/13! = 15*14 = 210
Manager
Joined: 12 Feb 2008
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Re: M11-29 [#permalink]  13 May 2008, 15:37
quote="ldpedroso"]I did with a different formula:
15!/13! = 15*14 = 210[/quote]

i did it in a similar way.
15*14 which is essentially 15!/13!
SVP
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Re: M11-29 [#permalink]  13 May 2008, 20:22
bsd_lover wrote:
Quite easy.

OA is C
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Current Student
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Re: M11-29 [#permalink]  16 May 2008, 07:48
I should elaborate where this formula n*n-1 is comin from

basically the formula is ..well first..you need to 2 chess players to play 1 game correct!

with this assumption in mind you are asking yourself ..the total number of games played..well that really means NC2..i.e in how many ways can you choose 2 players out of N..

NC2=N*(N-1)/2=total number of games played!

now in this question we are told number of games played=2 times each..so basically this becomes (n)(n-1)=15*14=210
SVP
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Schools: CBS, Kellogg
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Kudos [?]: 443 [0], given: 1

Re: M11-29 [#permalink]  16 May 2008, 19:10
Warm-hearted fresiha12! Thanks!
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Re: M11-29   [#permalink] 16 May 2008, 19:10
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# 15 chess players take part in a tournament. Every player

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