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There are 8 teams in a certain league and each team plays [#permalink]
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17 Jun 2012, 02:52
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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
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Last edited by Bunuel on 17 Jun 2012, 02:56, 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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Re: There are 8 teams in a certain league and each team plays [#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 \(C^2_{8}=28\). Answer: C. P.S. Please read and follow: rulesforpostingpleasereadthisbeforeposting133935.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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18 Nov 2012, 20:09
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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, 00:37
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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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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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16 May 2013, 03:33
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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, 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 [#permalink]
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There are 8 teams in a certain league and each team plays [#permalink]
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26 Nov 2014, 01:14
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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.htmlhope it helps.
Last edited by quentin.louviot on 13 Jan 2015, 06:51, 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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12 Jan 2015, 11:48
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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. 11, 22, 33 and then 21 (because you have 12), 32 (because you have 23). 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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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. 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 Final Answer: GMAT assassins aren't born, they're made, Rich
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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, 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.
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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20 May 2015, 02:10
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 onedayillYou'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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25 May 2016, 21:25
Attached is a visual that should help.
Attachments
Screen Shot 20160525 at 9.41.57 PM.png [ 75.15 KiB  Viewed 18142 times ]
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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16 Jun 2016, 05:00
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 (82)!] (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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27 Jun 2017, 05:43
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: Cheers, Brent
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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04 Jul 2017, 18:16
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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04 Jul 2017, 18:34
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 Sent from my SMN900 using GMAT Club Forum mobile app




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