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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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17 Jun 2012, 03:52
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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
[Reveal] Spoiler: OA

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

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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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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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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 $$C^2_{8}=28$$.

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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16 Nov 2012, 04:19
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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: 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.

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:
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
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

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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28 Dec 2012, 06:59
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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

$$=\frac{8!}{2!6!}=4*7 = 28$$

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

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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.

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

[Reveal] Spoiler:
C

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

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20 May 2015, 03:10
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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:

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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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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27 Dec 2013, 03:08
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There are 8 teams in a certain league and each team plays [#permalink]

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26 Nov 2014, 02:14
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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.

Last edited by quentin.louviot on 13 Jan 2015, 07:51, edited 2 times in total.

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

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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 $$C^2_{8}=28$$.

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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15 Nov 2012, 23:27
Hi Bunnel,

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

Sree

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

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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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17 May 2013, 21:18
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.

That's one of those tips that can make my life easier. I'm book marking this page. Thanks.
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Re: There are 8 teams in a certain league and each team plays [#permalink]

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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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12 Sep 2014, 06:36
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

total 8 teams & each game by 2 pair then 8C2

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

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26 Nov 2014, 02:57
If there are n teams need to play exactly once ,then they play with (n-1) teams but as they are playing together then, n(n-1)/2, which means nC2

So 8*7/2 =28.

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

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12 Jan 2015, 12:48
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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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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] 19 May 2015, 07:31

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