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There are n people competing in a chess tournament. Each com

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There are n people competing in a chess tournament. Each com [#permalink] New post 04 May 2017, 06:34
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Question Stats:

62% (00:00) correct 37% (02:57) wrong based on 8 sessions
There are n people competing in a chess tournament. Each competitor must play every other competitor k times. If n > 1 and k > 0, what is the total number of games played in the tournament?

A) kn – k
B) (n² – 2k)/2
C) k(n² – n)/2
D) (n² – 2nk + k)/2
E) (kn – 2k)/2
[Reveal] Spoiler: OA

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Brent Hanneson – Creator of greenlighttestprep.com
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1 KUDOS received
GRE Instructor
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Joined: 10 Apr 2015
Posts: 1246
Followers: 46

Kudos [?]: 1137 [1] , given: 7

Re: There are n people competing in a chess tournament. Each com [#permalink] New post 06 May 2017, 05:41
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GreenlightTestPrep wrote:
There are n people competing in a chess tournament. Each competitor must play every other competitor k times. If n > 1 and k > 0, what is the total number of games played in the tournament?

A) kn – k
B) (n² – 2k)/2
C) k(n² – n)/2
D) (n² – 2nk + k)/2
E) (kn – 2k)/2


Here's an approach that doesn't require any formal counting techniques:

Let's say a MATCH is when two competitors sit down to play their k games against each other.

If we ask each of the n competitors, "How many MATCHES did you have?", the answer will be n-1, since each competitor plays every other competitor, but does not play against him/herself.

So, n(n-1) = the total number of MATCHES

IMPORTANT: There's some duplication here.
For example, when Competitor A says that he/she played n-1 other competitors, this includes the match played against Competitor B. Likewise, when Competitor B says he/she played n-1 other competitors, this includes the match played against Competitor A.

So, in our calculation of n(n-1) = the total number of MATCHES, we included the A vs B match twice.
In fact, we counted every match two times.

To account for this duplication, we'll take n(n-1) and divide by 2 to get n(n-1)/2 MATCHES.

Since each match consists of k games, the total number of games = kn(n-1)/2

Check the answer choices....not there!
However, we can take kn(n-1)/2 and rewrite it as k(n² - n)/2

Answer:
[Reveal] Spoiler:
C


Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com
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Re: There are n people competing in a chess tournament. Each com   [#permalink] 06 May 2017, 05:41
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