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

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

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Updated on: 26 Feb 2019, 03:29
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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, 02:52.
Last edited by Bunuel on 26 Feb 2019, 03:29, edited 2 times in total.
Renamed the topic.
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Re: There are 8 teams in a certain league and each team plays each of the  [#permalink]

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16 Nov 2012, 03:19
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27
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 $$C^2_{8}=28$$.

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:
http://gmatclub.com/forum/how-many-diag ... 01540.html
http://gmatclub.com/forum/if-10-persons ... 10622.html
http://gmatclub.com/forum/how-many-diff ... 29992.html
http://gmatclub.com/forum/15-chess-play ... 55939.html
http://gmatclub.com/forum/there-are-5-c ... 27235.html
http://gmatclub.com/forum/if-each-parti ... 42222.html

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

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31 Jan 2013, 00:37
7
5
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 each of the  [#permalink]

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12 Jan 2015, 13:46
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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.

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 each of the  [#permalink]

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18 Nov 2012, 20:09
6
2
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.

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

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17 Jun 2012, 02:57
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5
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 $$C^2_{8}=28$$.

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

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Updated on: 13 Jan 2015, 06:51
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1
1
SreeViji wrote:
Hi Bunnel,

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

Sree

Hey SreeViji,

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, 01:14.
Last edited by quentin.louviot on 13 Jan 2015, 06:51, edited 2 times in total.
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Re: There are 8 teams in a certain league and each team plays each of the  [#permalink]

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27 Dec 2013, 02:08
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Re: There are 8 teams in a certain league and each team plays each of the  [#permalink]

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20 May 2015, 02:10
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.

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:

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 each of the  [#permalink]

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16 May 2013, 03:33
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10
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 each of the  [#permalink]

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25 May 2016, 21:25
1
Attached is a visual that should help.
Attachments

Screen Shot 2016-05-25 at 9.41.57 PM.png [ 75.15 KiB | Viewed 112796 times ]

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

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16 Jun 2016, 05:00
1
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

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

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19 Mar 2018, 10:00
1
dave13 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 $$C^2_{8}=28$$.

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

Bunuel you know I got confused by your shortcut solution until figured out all possible combinations in details . you know what surprises how this expression $$C^2_{8}=28$$
excludes the possibility of playing more than one game by two distinct teams, also it exludes repeated games like AB and BA....

let 8 teams be A, B, C, D, E, F, G, H

NUMBER OF GAMES PLAYES BY TWO TEAMS AS FOLLOWS:

AB BC CD DE EF
AC BD CE DF EG
AE BF CG DH
AF BG CH
AG BH
AH

Hi dave13

it is clearly mentioned in the question that each team plays against other team only Once. Hence you can safely use the formula provided by Bunuel. and if we say A plays against B then its same as saying B plays against A so order does not matter here. Hence there will be only one combination with A & B taken together and not two different combinations as stated by you AB & BA.
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Re: There are 8 teams in a certain league and each team plays each of the  [#permalink]

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14 Nov 2012, 05:23
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 $$C^2_{8}=28$$.

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 each of the  [#permalink]

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16 May 2013, 00:13
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 each of the  [#permalink]

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26 Dec 2013, 22:04
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 each of the  [#permalink]

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12 Jan 2015, 11:48
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 each of the  [#permalink]

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19 May 2015, 06:31
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.

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 each of the  [#permalink]

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Updated on: 06 Dec 2018, 10:46
Top Contributor
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

Cheers,
Brent
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Originally posted by GMATPrepNow on 27 Jun 2017, 05:43.
Last edited by GMATPrepNow on 06 Dec 2018, 10:46, edited 1 time in total.
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Re: There are 8 teams in a certain league and each team plays each of the  [#permalink]

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25 Aug 2017, 23:46
Try this watch this, similar examples-

Try pick up from Khan academy to built the foundation for permutations and combinations.

The questions are very interesting too.

Similar questions found in:

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Re: There are 8 teams in a certain league and each team plays each of the   [#permalink] 25 Aug 2017, 23:46

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