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

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Intern
Joined: 20 Nov 2013
Posts: 11
Location: Italy
GPA: 3.6
Re: 15 chess players take part in a tournament. Every player  [#permalink]

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09 Jan 2014, 04:21
formula for this type of problems.

1) If every team(player..) plays against others teams only ONCE. [n(n-1)]/2

2) If every team(player..) plays against others teams TWICE [n(n-1)]

where n is the number of teams..
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Joined: 02 Jul 2014
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15 chess players take part in a tournament. Every player  [#permalink]

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13 Jul 2014, 16:52
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

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Joined: 02 Sep 2009
Posts: 64939
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14 Jul 2014, 02:04
Intern
Joined: 02 Jul 2014
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14 Jul 2014, 02:37
Bunuel wrote:
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

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

Thank you Bunuel, you are using combination formula, I have understood your method, but I just wanted to understand the meaning of these terms in above post "15*15 - 15 = 210" and how to use same method for 3 matches. Thanks again.
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Posts: 64939
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14 Jul 2014, 02:41
mankul wrote:
Bunuel wrote:
mankul wrote:

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

Thank you Bunuel, you are using combination formula, I have understood your method, but I just wanted to understand the meaning of these terms in above post "15*15 - 15 = 210" and how to use same method for 3 matches. Thanks again.

Frankly I don't know what is the logic behind J's solution.
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15 chess players take part in a tournament. Every player  [#permalink]

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Updated on: 30 May 2020, 22:06
I didnt want to use the formulae so I tried a different approach.

Say we have 4 players, A, B, C, D
Any player cannot play against himself/herself

So we have a total of 4*4=16 total possibilities for 4 players but since they cant play against themselves we need to subtract 4 from 16 which gives us 16-4=12, as illustrated in the diagram below

-- A---B---C---D
A x---1---1---1
B 1---x---1---1
C 1---1---x---1
D 1---1---1---x

If you deploy this concept for 15 players, we have 15*15=225 total possibilities but since these 15 players cannot play against themselves, we subtract 15 from 225 which gives us 225-15=210

Please correct me if I am wrong with this approach.

Thanks!

Originally posted by himanshu11pathak on 30 May 2020, 19:44.
Last edited by himanshu11pathak on 30 May 2020, 22:06, edited 1 time in total.
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30 May 2020, 21:32
chess game is well played in indoor game and there are 210 game will be played and I am go with ans C
Re: 15 chess players take part in a tournament. Every player   [#permalink] 30 May 2020, 21:32

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