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Re: PT # 11 DS 9 [#permalink]
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nandinigaur wrote:
Bunuel wrote:
There are exactly 6 teams in league x. What was the total number of games played by the 6 teams last season?

(1) Each team in league x played each of the other teams at least once --> if each team played each of the other teams ONLY once then there was total of \(C^2_6=15\) games played (notice that each team played 5 games) but if each team played each of the other teams TWICE then there was total of \(2*C^2_6=30\) games played (notice that each team played 10 games). Not sufficient.

(2) No team in league x played more than 7 games. Clearly insufficient.

(1)+(2) If each team played each of the other teams ONLY once then there was total of \(C^2_6=15\) games played (each of the 6 teams played 5 games), but if after each team played each of the other teams ONLY once, some two teams played between each other once more then the total number of games played was 16 (4 teams played 5 games and 2 teams played 6 games). Not sufficient.

Answer: E.


Hi Bunnel

I get confused in at least and at most qs... here st 1 says that each team played atleast one game with the other team. I understand that this mean some could have played more than one.
but in the other qs about out of 2 atleast 1 snake is black and 1 snake is white... we understand it as 1 black snake...
please explain about qs with atleast atmost in the qs stem.


At least one means 1 or more;
At most 1 means 1 or less.

Nothing more in it.

I think the question you are talking about is this one: devil-s-dozen-129312-20.html#p1063888 But this question is different and has some other constrains.

P.S. If you have problems with most of the questions one of the reasons could be lack of fundamentals. So, I'd advice to go through the basics once more.
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Re: There are exactly 6 teams in league x. What was the total [#permalink]
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Dear Bunuel,

Can you explain "but if after each team played each of the other teams ONLY once, some two teams played between each other once more then the total number of games played was 16" what this means? Thanks
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Re: There are exactly 6 teams in league x. What was the total [#permalink]
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sri30kanth wrote:
Dear Bunuel,

Can you explain "but if after each team played each of the other teams ONLY once, some two teams played between each other once more then the total number of games played was 16" what this means? Thanks


Say each team played each of the other teams once: the number of games played = \(C^2_6=15\). After that, say two of the teams played one more game between each other. The total number of games played = 15 + 1 = 16.

Hope it's clear.
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Re: There are exactly 6 teams in league x. What was the total [#permalink]
Yeah let's assume two of the teams played one more game between each other but still the total no of games a team played would be less than 7. Isn't it? Or can't we guarantee that? Say A,B,C,D,E,F are six teams. After initial 15 games where each team played 5 times, is there a possibility of other teams playing randomly? Please clarify. Thank you
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Re: There are exactly 6 teams in league x. What was the total [#permalink]
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sri30kanth wrote:
Yeah let's assume two of the teams played one more game between each other but still the total no of games a team played would be less than 7. Isn't it? Or can't we guarantee that? Say A,B,C,D,E,F are six teams. After initial 15 games where each team played 5 times, is there a possibility of other teams playing randomly? Please clarify. Thank you


Not sure I understand your question...

Each team played each of the other teams once, then two team played one more game between each other (say that was the final game). Total games = 16. Each team played less than 7 games. What's unclear here?
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Re: There are exactly 6 teams in league x. What was the total [#permalink]
Bunuel wrote:
There are exactly 6 teams in league x. What was the total number of games played by the 6 teams last season?

(1) Each team in league x played each of the other teams at least once --> if each team played each of the other teams ONLY once then there was total of \(C^2_6=15\) games played (notice that each team played 5 games) but if each team played each of the other teams TWICE then there was total of \(2*C^2_6=30\) games played (notice that each team played 10 games). Not sufficient.


Can you please help me understand the logic behind this combination application "\(C^2_6=15\)" used when each team plays the other only once?

Thanks.
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Re: There are exactly 6 teams in league x. What was the total [#permalink]
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TheMastermind wrote:
Bunuel wrote:
There are exactly 6 teams in league x. What was the total number of games played by the 6 teams last season?

(1) Each team in league x played each of the other teams at least once --> if each team played each of the other teams ONLY once then there was total of \(C^2_6=15\) games played (notice that each team played 5 games) but if each team played each of the other teams TWICE then there was total of \(2*C^2_6=30\) games played (notice that each team played 10 games). Not sufficient.


Can you please help me understand the logic behind this combination application "\(C^2_6=15\)" used when each team plays the other only once?

Thanks.


The total number of games possible will be total number of different pairs possible out of 6 teams (one game per one pair), so \(C^2_6=15\).
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Re: There are exactly 6 teams in league x. What was the total [#permalink]
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Bunuel wrote:
TheMastermind wrote:
Bunuel wrote:
There are exactly 6 teams in league x. What was the total number of games played by the 6 teams last season?

(1) Each team in league x played each of the other teams at least once --> if each team played each of the other teams ONLY once then there was total of \(C^2_6=15\) games played (notice that each team played 5 games) but if each team played each of the other teams TWICE then there was total of \(2*C^2_6=30\) games played (notice that each team played 10 games). Not sufficient.


Can you please help me understand the logic behind this combination application "\(C^2_6=15\)" used when each team plays the other only once?

Thanks.


The total number of games possible will be total number of different pairs possible out of 6 teams (one game per one pair), so \(C^2_6=15\).


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Re: There are exactly 6 teams in league x. What was the total [#permalink]
Bunuel wrote:

The total number of games possible will be total number of different pairs possible out of 6 teams (one game per one pair), so \(C^2_6=15\).


Hi, I am little confused about the use of the formula because according to the GMAT club Math book, the formula for combinations is \(C^n_k=\) n! / k! (n-k)!
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Re: There are exactly 6 teams in league x. What was the total [#permalink]
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ameyaprabhu wrote:
Bunuel wrote:

The total number of games possible will be total number of different pairs possible out of 6 teams (one game per one pair), so \(C^2_6=15\).


Hi, I am little confused about the use of the formula because according to the GMAT club Math book, the formula for combinations is \(C^n_k=\) n! / k! (n-k)!


6C2, \(C^2_6\), \(C^6_2\) all mean the same thing - choosing different groups of 2 out of 6: \(\frac{6!}{2!4!}\). Can it be anything else? Can we choose 6 out of 2?
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Re: There are exactly 6 teams in league x. What was the total [#permalink]
Dear Bunuel,
Could you pls clarify -- using what formula did you calculated number of played games?
"(1) Each team in league x played each of the other teams at least once --> notice that each team played 5 games"
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Re: There are exactly 6 teams in league x. What was the total [#permalink]
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stepanyan wrote:
Dear Bunuel,
Could you pls clarify -- using what formula did you calculated number of played games?
"(1) Each team in league x played each of the other teams at least once --> notice that each team played 5 games"


The total number of teams is six. If each team played every other team exactly once, then the total number of games would equate to the number of pairs that can be formed from these six teams. Mathematically, this is represented as 6C2, which results in 15 total games. When we look at the number of games each team played, it's five. This is because, out of the six teams, each has five potential opponents.
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Re: There are exactly 6 teams in league x. What was the total [#permalink]
Bunuel,
thanks a lot!
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Re: There are exactly 6 teams in league x. What was the total [#permalink]
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