There are 5 chess amateurs playing in Villa's chess club

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There are 5 chess amateurs playing in Villa's chess club [#permalink]

07 Feb 2012, 21:36

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63% (01:36) correct

36% (00:16) wrong

based on 2 sessions

There are 5 chess amateurs playing in Villa's chess club tournament. If each chess amateur plays with exactly 4 other amateurs, what is the total number of chess games possible to be played in the tournament?

A 10
B 20
C 40
D 60
E 120

[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Feb 2012, 02:51, edited 1 time in total.

Edited the question and added the OA

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Re: Combinations [#permalink]

07 Feb 2012, 22:15

Smita04 wrote:

There are different ways of approaching this question.

Method 1:
Take the first amateur. He plays a game with each of the other four i.e. 4 games.
Now take the second one. He has already played a game with the first one. He plays 3 games with the rest of the 3 amateurs i.e. 3 more games are played.
Now take the third amateur. He has already played a game each with the first and the second amateur. Now he plays 2 games with the remaining 2 amateurs so 2 more games are played.
Now go on to the fourth amateur. He has already played 3 games with the first 3 amateurs. He just needs to play a game with the last one i.e. 1 more game is played.
The last amateur has already played 4 games.
Total no of games = 4+3+2+1 = 10

Method 2:
Each person is one participant of 4 games. So there are in all 4*5 = 20 instances of one participant games. But each game has 2 participants so total number of games = 20/2 = 10
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Re: Combinations [#permalink]

08 Feb 2012, 02:51

Smita04 wrote:

The easiest way probably would be to realize that total # of games possible will be total # of different pairs possible out of 5 players (one game per one pair): C^2_5=\frac{5!}{3!*2!}.

Answer: A.

Check similar questions to practice:
m07-81505.html
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Hope it helps.
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Re: There are 5 chess amateurs playing in Villa's chess club [#permalink]

08 Feb 2012, 06:16

First player = 4 different games possible
Second player = 3 different games possible (1 has already counted)
third player = 2 different games possible
fourht player = 1 different game possible

--> 10 Games or Answer A

Re: There are 5 chess amateurs playing in Villa's chess club [#permalink] 08 Feb 2012, 06:16

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There are 5 chess amateurs playing in Villa's chess club

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