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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:

61% (00:50) correct 38% (02:44) wrong based on 18 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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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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Re: There are n people competing in a chess tournament. Each com [#permalink] New post 17 Dec 2018, 11:30
Only one answer is correct and it will be correct for all n and for all k, so we can make up a number and check each.

Suppose n = 5 and each play k=2 games.

Lets label the 5 players like so

P1 P2 P3 P4 P5

Player 1 will be able to play 4 other people, and play each k = 2 games so
# games played by P1 = 4 *2 = 8
Player 2 will be able to play the other 3 people (he can't go back and play player 1 that will be double counting), and he plays each two times.
# games played by P2 = 3 *2 = 6
Player 3 will be able to play the other 2 people (he can't go back and play player 1 or 2 that will be double counting), and he plays each two times.
# games played by P2 = 2 *2 = 4
Player 4 will be able to play the other 1 person (he can't go back and play player 1, 2, or 3, that will be double counting), and he plays him/her two times.
# games played by P2 = 1 *2 = 2

Total number of games played: 20

Plug in k = 2 and n = 5, only choice C gets you the correct answer.

But that is too many players, we could have chosen n = 2 and k = 1, thus leaving us with only 2 matches possible between P1 and P2.

Choice C satisfies this and the others do not.
Re: There are n people competing in a chess tournament. Each com   [#permalink] 17 Dec 2018, 11:30
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