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# There are n people competing in a chess tournament. Each competitor...

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There are n people competing in a chess tournament. Each competitor... [#permalink]

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30 Apr 2017, 09:20
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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

*kudos for all correct solutions
[Reveal] Spoiler: OA

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Brent Hanneson – Founder of gmatprepnow.com

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Joined: 13 Apr 2015
Posts: 1600
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Concentration: Strategy, General Management
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Re: There are n people competing in a chess tournament. Each competitor... [#permalink]

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30 Apr 2017, 09:39
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If n people play with every other member exactly once (k = 1) then the total number of matches = nC2 = (n(n- 1))/2

If they have to play with each other k times then the total = k(n^2 - n)/2

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Posts: 874
Re: There are n people competing in a chess tournament. Each competitor... [#permalink]

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30 Apr 2017, 10:58
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GMATPrepNow 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

*kudos for all correct solutions

• If each player plays with each other only once then the total number games is simply given by
o $$^nC_2 = \frac{n!}{[2!(n-2)!]} = \frac{[n(n-1)]}{2}$$
o Where n is the number of players and we choose 2 players at a time to play a single game.
• But then each player is playing its competitor “k” times.
o Hence the total number of games would be multiplied “k” times.
o Thus the total number of games = k * Total number of games played once $$= k*n*\frac{(n-1)}{2} = k*\frac{(n^2 – n)}{6}$$
• Hence the correct answer is Option C.

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Posts: 2150
Re: There are n people competing in a chess tournament. Each competitor... [#permalink]

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01 May 2017, 06:18
Expert's post
Top Contributor
These are great solutions. There are still a couple of completely different ways to solve this question.
Any takers?

Cheers,
Brent
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Joined: 26 Mar 2013
Posts: 1506
Re: There are n people competing in a chess tournament. Each competitor... [#permalink]

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01 May 2017, 06:21
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GMATPrepNow 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

I will do this logically

If I choose 3 teams to meet only 1 round. then total number of games =3

n= 3 & k=1 & total =3

Plug in the answer choices

A) kn – k ..................................3-1=2.............................Eliminate

B) (n² – 2k)/2......................... (9-2)/2.............................Eliminate

C) k(n² – n)/2...........................1 * (9-3)/2=3....................Keep

D) (n² – 2nk + k)/2..................(9-6+1)/2.........................Eliminate

E) (kn – 2k)/2..........................(3-2)/2.............................Eliminate

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Posts: 874
Re: There are n people competing in a chess tournament. Each competitor... [#permalink]

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01 May 2017, 06:41
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Expert's post
GMATPrepNow 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

*kudos for all correct solutions

Let me assume a student who does not know how to use permutation and combination and yet he wants to solve this questions.

He can use a bit of logic and hit and trial to get the answer.

So suppose if I consider, N = 2 and k = 2, the number of games played should be 2 ( easy to visualize, A and B are two people, they play matches 2 times)

A) kn – k = 4 - 2 = 2
B) (n² – 2k)/2 = 4 -4/2 = 0 = Not possible
C) k(n² – n)/2 = 2.2/2 = 2
D) (n² – 2nk + k)/2 = 4 - 8 +2/2 = negative = Not Possilbe
E) (kn – 2k)/2 = 4- 4/2 = 0 Not possible

Now we are left with, Option A and C.

Take N = 3 and k = 1, if there are three people A, B, C, the matches they can play are AB, BC and AC i.e. three matches.

A. 1.3 - 1 = 2 Not possible
C. 1.(9-3)2/ = 3 Hence C has to be the answer.

Thanks,
Saquib
Quant Expert
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Joined: 12 Sep 2015
Posts: 2150
Re: There are n people competing in a chess tournament. Each competitor... [#permalink]

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01 May 2017, 06:58
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GMATPrepNow 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

[Reveal] Spoiler:
C

Cheers,
Brent
_________________

Brent Hanneson – Founder of gmatprepnow.com

Intern
Joined: 05 Jun 2014
Posts: 8
Re: There are n people competing in a chess tournament. Each competitor... [#permalink]

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25 Oct 2017, 23:49
EgmatQuantExpert wrote:
GMATPrepNow 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

*kudos for all correct solutions

• If each player plays with each other only once then the total number games is simply given by
o $$^nC_2 = \frac{n!}{[2!(n-2)!]} = \frac{[n(n-1)]}{2}$$
o Where n is the number of players and we choose 2 players at a time to play a single game.
• But then each player is playing its competitor “k” times.
o Hence the total number of games would be multiplied “k” times.
o Thus the total number of games = k * Total number of games played once $$= k*n*\frac{(n-1)}{2} = k*\frac{(n^2 – n)}{6}$$
• Hence the correct answer is Option C.

shouldn't the answer be $$k*\frac{(n^2 – n)}{2}$$ and not $$k*\frac{(n^2 – n)}{6}$$
I assume this to be a typo
Re: There are n people competing in a chess tournament. Each competitor...   [#permalink] 25 Oct 2017, 23:49
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# There are n people competing in a chess tournament. Each competitor...

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