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15 chess players take part in a tournament. Every player

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SVP
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15 chess players take part in a tournament. Every player [#permalink]

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New post 13 May 2008, 00:59
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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Re: M11-29 [#permalink]

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New post 13 May 2008, 01:09
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!

Kudos [?]: 668 [0], given: 210

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Re: M11-29 [#permalink]

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New post 13 May 2008, 06:38
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

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Re: M11-29 [#permalink]

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New post 13 May 2008, 11:28
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..

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Re: M11-29 [#permalink]

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New post 13 May 2008, 16:04
I did with a different formula:
15!/13! = 15*14 = 210

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Re: M11-29 [#permalink]

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New post 13 May 2008, 16:37
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!

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Re: M11-29 [#permalink]

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New post 13 May 2008, 21:22
bsd_lover wrote:
Quite easy.


:devil

OA is C
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Re: M11-29 [#permalink]

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New post 16 May 2008, 08:48
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

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Re: M11-29 [#permalink]

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New post 16 May 2008, 20:10
Warm-hearted fresiha12! Thanks!
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Re: M11-29   [#permalink] 16 May 2008, 20:10
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15 chess players take part in a tournament. Every player

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