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

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CEO
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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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55% (hard)

Question Stats:

63% (01:42) correct 37% (03:11) wrong based on 154 sessions

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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

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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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1
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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Re: There are n people competing in a chess tournament. Each competitor...  [#permalink]

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30 Apr 2017, 10:58
2
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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Re: There are n people competing in a chess tournament. Each competitor...  [#permalink]

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01 May 2017, 06:18
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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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Re: There are n people competing in a chess tournament. Each competitor...  [#permalink]

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01 May 2017, 06:21
1
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

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

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01 May 2017, 06:41
1
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
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CEO
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Posts: 2713
Re: There are n people competing in a chess tournament. Each competitor...  [#permalink]

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01 May 2017, 06:58
1
Top Contributor
1
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

However, we can take kn(n-1)/2 and rewrite it as k(n² - n)/2

Cheers,
Brent
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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... &nbs [#permalink] 25 Oct 2017, 23:49
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