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There are 8 teams in a certain league and each team plays

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There are 8 teams in a certain league and each team plays  [#permalink]

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New post Updated on: 17 Jun 2012, 03:56
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There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A. 15
B. 16
C. 28
D. 56
E. 64

Originally posted by sarb on 17 Jun 2012, 03:52.
Last edited by Bunuel on 17 Jun 2012, 03:56, edited 1 time in total.
Edited the question and added the OA.
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There are 8 teams in a certain league and each team plays  [#permalink]

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New post 16 Nov 2012, 04:19
3
17
Sachin9 wrote:
Bunuel wrote:
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A. 15
B. 16
C. 28
D. 56
E. 64


The total # of games played would be equal to the # of different pairs possible from 8 teams, which is \(C^2_{8}=28\).

Answer: C.

P.S. Please read and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention ot the points #3 and #8.

I would like to learn about \(C^2_{8}=28\). Manhattan Book doesn't discuss this approach. They have anagram approach.


Well, the game is played by 2 teams. How many games are needed if there are 8 teams and each team plays each of the other teams exactly once? The number of games will be equal to the number of different pairs of 2 teams we can form out of 8 teams (one game per pair). How else?

Similar questions to practice:
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 31 Jan 2013, 01:37
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it is not easy and is harder if we are on the test date.

there are 8 team
to take out 2 teams, IF ORDER MATTERS we have 8*7
but in fact order does not matter

8*7/2=28

princeton gmat book explain this point wonderfully.
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 17 Jun 2012, 03:57
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sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A. 15
B. 16
C. 28
D. 56
E. 64


The total # of games played would be equal to the # of different pairs possible from 8 teams, which is \(C^2_{8}=28\).

Answer: C.

P.S. Please read and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention ot the points #3 and #8.
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 14 Nov 2012, 06:23
Bunuel wrote:
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A. 15
B. 16
C. 28
D. 56
E. 64


The total # of games played would be equal to the # of different pairs possible from 8 teams, which is \(C^2_{8}=28\).

Answer: C.

P.S. Please read and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention ot the points #3 and #8.

I would like to learn about \(C^2_{8}=28\). Manhattan Book doesn't discuss this approach. They have anagram approach.
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 18 Nov 2012, 21:09
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These type of problems can be solved with a simple diagram.

1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.


I tried uploading the diagram but unsuccessful.
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 16 May 2013, 01:13
Bunuel wrote:

The total # of games played would be equal to the # of different pairs possible from 8 teams, which is \(C^2_{8}=28\).


Hi Bunuel,

I have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula?

Thanks a lot.
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 16 May 2013, 04:33
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mywaytomba wrote:
Bunuel wrote:

The total # of games played would be equal to the # of different pairs possible from 8 teams, which is \(C^2_{8}=28\).


Hi Bunuel,

I have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula?

Thanks a lot.


Check combinatorics chapter of Math Book for theory: math-combinatorics-87345.html

Also check some questions on combinations to practice:
DS: search.php?search_id=tag&tag_id=31
PS: search.php?search_id=tag&tag_id=52

Hard questions on combinations and probability with detailed solutions: hardest-area-questions-probability-and-combinations-101361.html (there are some about permutation too)

Hope it helps.
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 26 Dec 2013, 23:04
Lets assume the question asks There are 8 teams in a certain league and each team
plays each of the other teams exactly twice. If each
game is played by 2 teams, what is the total number
of games played?

Then is 28*2 the correct approach?
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 27 Dec 2013, 03:08
2
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theGame001 wrote:
Lets assume the question asks There are 8 teams in a certain league and each team
plays each of the other teams exactly twice. If each
game is played by 2 teams, what is the total number
of games played?

Then is 28*2 the correct approach?


Yes. Check here: 15-chess-players-take-part-in-a-tournament-every-player-55939.html and here: there-are-8-teams-in-a-certain-league-and-each-team-plays-132366.html

Similar questions to practice:
how-many-diagonals-does-a-polygon-with-21-sides-have-if-one-101540.html
if-10-persons-meet-at-a-reunion-and-each-person-shakes-hands-110622.html
10-business-executives-and-7-chairmen-meet-at-a-conference-126163.html
how-many-different-handshakes-are-possible-if-six-girls-129992.html
15-chess-players-take-part-in-a-tournament-every-player-55939.html
there-are-5-chess-amateurs-playing-in-villa-s-chess-club-127235.html
if-each-participant-of-a-chess-tournament-plays-exactly-one-142222.html
there-are-8-teams-in-a-certain-league-and-each-team-plays-134582.html
there-are-8-teams-in-a-certain-league-and-each-team-plays-132366.html
in-a-kickball-competition-of-9-teams-how-many-possible-ma-161846.html

Hope it helps.
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There are 8 teams in a certain league and each team plays  [#permalink]

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New post Updated on: 13 Jan 2015, 07:51
2
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SreeViji wrote:
Hi Bunnel,

I would also like to learn this approach. Can u help me?

Sree



Hey SreeViji,

I think i have something to help you.
The answer here is the combination 8C2 (8 teams Choose 2) which mean \frac{8!}{6!x2!} --> \frac{8x7}{2}

To understand that we just have to think that each of the 8 team plays against 7 other (8x7) but they play each team exactly once so we divide the total by 2.

We divide by 2 because "TEAM A VS TEAM B" is the same as "TEAM B VS TEAM A"

So we end up with \frac{8x7}{2}= 28

If you still have trouble with combination and permutation check out this website it's well done,
http://www.mathsisfun.com/combinatorics ... tions.html

hope it helps.

Originally posted by quentin.louviot on 26 Nov 2014, 02:14.
Last edited by quentin.louviot on 13 Jan 2015, 07:51, edited 2 times in total.
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There are 8 teams in a certain league and each team plays  [#permalink]

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New post 12 Jan 2015, 12:48
1
I also used a table to do this, like that:

1_2_3_4_5_6_7_8
1_1_1_1_1_1_1_1
2_2_2_2_2_2_2_2
3_3_3_3_3_3_3_3
4_4_4_4_4_4_4_4
5_5_5_5_5_5_5_5
6_6_6_6_6_6_6_6
7_7_7_7_7_7_7_7
8_8_8_8_8_8_8_8

Then you delete the same team pairs: e.g. 1-1, 2-2, 3-3 and then 2-1 (because you have 1-2), 3-2 (because you have 2-3). After you cross out the first 2 columns you then see that you cross out everything from the diagonal and below. The remaining is 28.

However, the 8!/2!*6! approach is better, because if you have many numbers the table will take forever to draw. In case there is sth similar though and your brain gets stuck, use the table...
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 12 Jan 2015, 14:46
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Hi All,

Using the Combination Formula IS one way to approach these types of questions, but it's not the only way. Sometimes the easiest way to get to a solution on Test Day is to just draw a little picture and keep track of the possibilities....

Let's call the 8 teams: ABCD EFGH

We're told that each team plays each other team JUST ONCE.

Start with team A....
A plays BCD EFGH = 7 games total

Team B has ALREADY played team A, so those teams CANNOT play again...
B plays CD EFGH = 6 more games

Team C has ALREADY played teams A and B, so the following games are left...
C plays D EFGH = 5 more games

At this point, you should notice a pattern: the additional number of games played is reduced by 1 each time. So what we really have is:

7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 28 games played

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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 19 May 2015, 07:31
pranav123 wrote:
These type of problems can be solved with a simple diagram.

1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.


I tried uploading the diagram but unsuccessful.


Hi,

I don;t think we need to count the half spaces. with half space count is 36.
without half space - count: 28.
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 20 May 2015, 03:10
2
onedayill wrote:
pranav123 wrote:
These type of problems can be solved with a simple diagram.

1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.


I tried uploading the diagram but unsuccessful.


Hi,

I don;t think we need to count the half spaces. with half space count is 36.
without half space - count: 28.


Dear onedayill

You're right! :)

The boxes along the diagonal (these are the boxes that contribute to half spaces) represent a team playing with itself. Since that is not possible, these boxes should not be included in the counting.

I noticed in the thread above that a few students had doubts about the expression 8C2. If any of the current students too have such a doubt, here's how this question could be solved visually:

Image

There are 7 ways in which Team 1 can play with another team.
Similarly, there are 7 ways for each of the 8 teams to choose its playing opponent.

But it's easy to see that the red zone is essentially a duplication of the blue zone. For example, (Team 1 playing with Team 2) is the same case as (Team 2 playing with Team 1)

So, the correct answer will be: 8(that is, the number of teams)*7(that is, the number of ways in which each team can choose its playing opponent)/2 = 28

Hope this was useful!

Best Regards

Japinder
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 25 May 2016, 22:25
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Attached is a visual that should help.
Attachments

Screen Shot 2016-05-25 at 9.41.57 PM.png
Screen Shot 2016-05-25 at 9.41.57 PM.png [ 75.15 KiB | Viewed 47992 times ]


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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 16 Jun 2016, 06:00
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sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A. 15
B. 16
C. 28
D. 56
E. 64



We are given that there are 8 teams in a league and that each game is played by 2 teams. Note that each team does not play itself, and the order of pairing each team with its opponent doesn't matter. [For example, the pairing of (Team A vs. Team B) is identical to the pairing of (Team B vs. Team A).] The situation can therefore be solved by finding the number of combinations of 8 items taken 2 at a time, or 8C2, as follows:

8C2 = 8! / [2! x (8-2)!]

(8 x 7 x 6!) / (2! x 6!)

(8 x 7)/2!

(8 x 7)/ 2

4 x 7 = 28

Answer C
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 27 Jun 2017, 06:43
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sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A. 15
B. 16
C. 28
D. 56
E. 64


There are 8 teams. If we ask each team, "How many teams did you play?" we'll find that each team played 7 teams, which gives us a total of 56 games (since 8 x 7 = 56).

From here we need to recognize that each game has been COUNTED TWICE.
For example, if Team A and Team B play a game, then Team A counts it as a game, and Team B ALSO counts it as a game.

So, to account for the DUPLICATION, we'll divide 56 by 2 to get 28

Answer:

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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 04 Jul 2017, 19:16
Hi,

If this question was modified to state

1. 'each team played 3 games with every other team'
2. 'each team played a total of 3 games, not necessarily all 3 with the same team'

Then will the total number of matches be\(^8C_3?\)

Also, do 'number of combinations' and 'number of choices' both deploy the \(^nC_k?\) formula?
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Re: There are 8 teams in a certain league and each team plays  [#permalink]

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New post 04 Jul 2017, 19:34
There is one approach which is the quickest one:

If we ask each team how many games it played, each team will say 7. Hence total 7×8=56 games are expected. However, each game has been counted twice, thus 56/2 = 28

The other way is to ask each team how many matches it has played such that each match is unique. The answer will be 7+6+5+4+3+2+1 = 28

Hence the answer is : (C) 28

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Re: There are 8 teams in a certain league and each team plays &nbs [#permalink] 04 Jul 2017, 19:34

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