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There are 8 teams in a certain league and each team plays [#permalink]
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10 May 2012, 20:45
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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? A. 15 B. 16 C. 28 D. 56 E. 64
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Re: There are 8 teams in a certain league and each team plays... [#permalink]
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10 May 2012, 21:16
IMO 54 matches
Let's say first team plays with 7 other teams  resulting in 7 matches, taking this further in the similar way, the number of matches between 8 teams would be 7+6+5+4+3+2+1 = 28.. as in this case each team is playing with other twice, the no of matches would be 28*2 = 56
Last edited by gmihir on 11 May 2012, 00:13, edited 1 time in total.



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Re: There are 8 teams in a certain league and each team plays... [#permalink]
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10 May 2012, 21:19
Every team plays with 7 teams...so total no of matches = 8 x 7 = 56. Now, each match is played twice => 56 x 2 But 2 teams play a match => 56 x 2 /2 = 56.



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Re: There are 8 teams in a certain league and each team plays... [#permalink]
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10 May 2012, 21:31
Bunuel, can you please comment on this?



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Re: There are 8 teams in a certain league and each team plays [#permalink]
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10 May 2012, 21:37
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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13 Sep 2013, 10:11
But the official answer is C, and I got that using combination [8!/(2!6!)]. Did I miss anything? Please help.



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Re: There are 8 teams in a certain league and each team plays [#permalink]
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13 Sep 2013, 10:55



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Re: There are 8 teams in a certain league and each team plays [#permalink]
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14 Sep 2013, 09:44
Thanks Bunuel! I caught my mistake. I was looking for other ways of solving question 133 OG 13th edi (121 in 12th edi), and that is how I landed here. Similar version of this question in the 13th and 12th edi OGs has the word 'once' instead of 'twice', as in the above question. My bad !! Is my approach at solving the question correct?



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Re: There are 8 teams in a certain league and each team plays [#permalink]
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14 Sep 2013, 09:55



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Re: There are 8 teams in a certain league and each team plays [#permalink]
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30 Jul 2014, 02:56
You have 8 teams, hence each each match day there are 4 games (since is 1v1 = 2 teams per match).
Every team has 7*2 (plays each game twice) match days, hence 4*14 games which is 56.
Answer D.



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Re: There are 8 teams in a certain league and each team plays [#permalink]
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23 Aug 2014, 10:11
Bunuel wrote: Smita04 wrote: 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?
A. 15 B. 16 C. 28 D. 56 E. 64 # of different pairs possible from 8 teams is \(C^2_{8}=28\), since each pair plays twice between each other than total # of games is 2*28=56. Answer: D. I'm a little confused here  why are we using the combination formula and NOT the permutation formula. We don't really care for these teams to be arranged alphabetically. Similar to if the letters are to be arranged alphabetically, meaning, ab, ac, ad, bc, bd, then we would use combination. But we don't care if team D plays B vs. team B playing team D. Since order is NOT important, wouldn't we use permutation. There were a few similar problems: 1) How many 2 letters words can be made out of ABCD and in alphabetical order  2C4 = 6 2) How many unique 4 letter words can be made from 10 letters but ABCDE and EDCBA are considered different = 10P4 = 10!/6! Doesn't this question fall into the Permutation area?



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Re: There are 8 teams in a certain league and each team plays [#permalink]
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18 Sep 2015, 09:47
Bunuel wrote: Smita04 wrote: 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?
A. 15 B. 16 C. 28 D. 56 E. 64 # of different pairs possible from 8 teams is \(C^2_{8}=28\), since each pair plays twice between each other than total # of games is 2*28=56. Answer: D. Can you show the calculation for 28? I am confused for the formula of combination. Shouldn't it be 8C2? Any possible theory on this from GMATCLUB?
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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19 Sep 2015, 06:03
harishbiyani8888 wrote: Bunuel wrote: Smita04 wrote: 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?
A. 15 B. 16 C. 28 D. 56 E. 64 # of different pairs possible from 8 teams is \(C^2_{8}=28\), since each pair plays twice between each other than total # of games is 2*28=56. Answer: D. Can you show the calculation for 28? I am confused for the formula of combination. Shouldn't it be 8C2? Any possible theory on this from GMATCLUB? It is 8C2. It is thee number of ways in which 2 teams can be selected out of 8. \(nCr = \frac{n!}{(nr)!(r!)\) \(8C2 = \frac{8!}{(82)!(2!)\) = \(\frac{8*7*6!}{6!*2!}\) = \(\frac{8*7}{2*1}\) =28 These are the total number of matches 8 teams can play when each team plays 1 match against all the other teams. Each team plays 2 matches. So total number of matches each team plays is 28*2 = 56



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Re: There are 8 teams in a certain league and each team plays [#permalink]
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19 Sep 2015, 16:01
Hi All, This prompt is remarkably similar to the following prompt: thereare8teamsinacertainleagueandeachteamplays134582.htmlHowever, since each team plays each other team TWICE (and not just once), the total number of games played is doubled. 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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Re: There are 8 teams in a certain league and each team plays [#permalink]
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08 May 2017, 06:40
If only one match is to be played : Total number of match one team will play : 7 (with each of the 7 other team) Hence all 8 teams will play : 7 x 8 Matches = 56 matches in all However, this contains repeated matches for each team .. Hence, in actual total matches will be = 56/2 = 28 So if 2 matches are to be played : 28 x 2 = 56 total matches OPtion C
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