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In a certain play, each of four players must play with each [#permalink]
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04 Jun 2009, 10:38
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In a certain play, each of four players must play with each of other players exactly once. If no game ended in a tie, did someone win one and lose two games? 1). No one won all the games. 2). No one lost all the games.



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Re: play [#permalink]
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04 Jun 2009, 13:35
Number of games =6 1). No one won all the games. Suppose A won all the games. ABA ACA ADA BC BD CD In any case the required condition “did someone win one and lose two games” will not be met. So (1) is SUFFICIENT. 2). No one lost all the games. EX1. Suppose A lost all the games. ABB ACC ADD BCB BD CDD So C won 1 game and lost 2 EX2. Suppose A lost all the games. ABB ACC ADD BCC BD CDD So C won 2 game and lost 1 So (1) is INSUFFICIENT. IMO A
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Re: play [#permalink]
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04 Jun 2009, 16:07
vcbabu wrote: In a certain play, each of four players must play with each of other players exactly once. If no game ended in a tie, did someone win one and lose two games? 1). No one won all the games. 2). No one lost all the games. There are 4x3 games, then dividing by 2 to eliminate repeats ie (player 1 w/ 2 is the same as 2 w/ 1). So 6. In total we have 6 wins, 6 losses, and each person gets 3. Nobody won all the games > everyone gets >=1 loss. So we have 2 losses left. A LLL B LWW C LWW D LWW 2 Sufficient After playing around with it Will post a more detailed solution
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Re: play [#permalink]
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04 Jun 2009, 16:10
waiting for Hades's explanation . BTW whats OA?
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Re: play [#permalink]
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04 Jun 2009, 22:16
It's a network flow. I set it up, very basic. I tried to solve it mentally but no dice, there is no configuration that works. Let me explain it another way We're going to have 6 matches, so 6 wins & 6 losses. Let's try and make a case where nobody Wins 1 game & Loses 2 games. If nobody can have 3 losses nor 2 losses (from condition in (2), and from additional condition we impose), then someone can have at most 1 loss. But we need to distribute 6 losses total...so let's give everyone 1 loss, and we have 2 losses left. But where do we put those? If we put another loss onto 2 different people, we violate our own condition (no 2 losses). And if we try to stick the 2 losses onto one player, we violate the condition in (2), which states that no player can lose all 3. Hence there will always be at least 1 player who wins 1 but loses 2. B
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Re: play [#permalink]
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04 Jun 2009, 23:31
More graphically: We're going to have \(6\) wins and \(6\) losses. Let's represent the players via \(A/B/C/D\). \(L\) = loss \(W\) = win Condition in (2): No player can have 3 losses[/m].
Let's look at the question: [b]Can a player have 2 losses?The YES case means a player CAN have 2 losses. Let's see if we can make a NO case, that is no player has 2 losses. If it's impossible to make a NO case, then the answer is always YES So we have 6 losses: \(LLLLLL\) & 4 players: \(A ???\) \(B ???\) \(C ???\) \(D ???\) So since nobody can have 3 losses, or 2 losses, let's give everyone 1 loss: \(A L??\) \(B L??\) \(C L??\) \(D L??\) We have 2 L's left. Where do we put them? If we put the 2 losses onto 1 player, we have 3 losses, which violates our condition that no player has 3 losses. But our only other choice is to put one loss onto 2 players which violates our assumption for the NO case that no player has 2 losses. Hence the NO case is impossible. Sufficient
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Re: play [#permalink]
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09 Jun 2009, 01:09
From Hade's approach, it clearly that We have 4 people A,B,C,D and 6 games as follow: AB AC AD BC BD DC See that we have total 6 Lose and 6 Win, each person play 3 games with other people. 1) If no one wills all, at most a person has 1 lose. So remains 2 lose. These 2 loses can be distributed to one A,B,C or D to make a person who has 3 all loses and each other has 1 lose 2 wins; or they can be distributed to two other people to make 2 people who have 2 loses 1 win. So insuf 2) Because no one lose all 3 games, let's distribute 6 loses into 3 people have 2 loses each; or 2 people have 2 loses each and 2 people have 1 lose each. So suf Hence B










