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# A certain league has four divisions. The respective

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Manager
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A certain league has four divisions. The respective [#permalink]

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10 Jan 2008, 16:27
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A certain league has four divisions. The respective divisions had 9, 10, 11, and 12 teams qualify for the playoffs. Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games -- in order to determine its champion. The four division champions then played in a single-elimination tournament -- where a team is eliminated upon losing one game -- in order to determine the overall league champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?

(A) 79
(B) 83
(C) 85
(D) 87
(E) 88
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Re: Games league champion A3 [#permalink]

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11 Jan 2008, 04:56
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Each game played will produce 1 win and 1 loss. The easiest way I can think of is to count the losses. For example, if there are 9 teams playing and it's double elimination then 18 losses would eliminate everyone. However, we want 1 team remaining so we're looking for 17 losses instead.

For X teams they'll play (2x-1) games in double elimination and (x-1) for single elimination.

9 Team League: We can have up to 17 losses
10 Team League: We can have up to 19 losses
11 Team League: We can have up to 21 losses
12 Team League: We can have up to 23 losses

Example:
1234 plays 5678 (one loss for 1234)
1234 plays 5678 (one loss apiece)
1234 plays 5678 (1234 are eliminated)
56 plays 78 (56 are eliminated)
7 plays 8 (8 is eliminated)
7 plays 9 (one loss apiece)
7 plays 9 (9 wins it all)

Now we go into the single elimination tournament with (17+19+21+23=80) games and 4 teams remaining

12 plays 34 (34 wins)
3 plays 4 (4 wins it all)

3 games in the final championship

80+3 =83 games

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Re: Games league champion A3 [#permalink]

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11 Jan 2008, 15:06
I may be wrong but here is my logic..
its true that in 9 teams there will have 17 losses but each team will play twice so total games played will be 18
so on division level, we will have
9*2=18
10*2=20
11*2=22
12*2=24
total = 18+20+22+24 = 84

After this four team will play total of 3 games to choose the winner
so, total games will be 84+3=87
There is high chance that my answer will be wrong but this is what i would choose in the test.
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Re: Games league champion A3 [#permalink]

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11 Jan 2008, 15:21
Example from my earlier post about the 9 team league

Quote:
1234 plays 5678 (one loss for 1234) (4 games)
1234 plays 5678 (one loss apiece) (4 games)
1234 plays 5678 (1234 are eliminated) (4 games)
56 plays 78 (56 are eliminated) (2 games)
7 plays 8 (8 is eliminated) (1 game)
7 plays 9 (one loss apiece) (1 game)
7 plays 9 (9 wins it all) (1 game)

4+4+4+2+1+1+1 = 17 games

As you can see it's not as simple as "each team plays 2 games". Team 7 in the example actually played 7 games! and no team up there played fewer than 3 games (except for team 9, but that's because they came in right at the end).

And even if each team were to only play 2 games, you can't just multiply the number of teams by 2 because each game involves 2 teams and if you count each of them you'll be counting each game twice.

Here's an example with 10 teams:

Quote:
01234 plays 56789 (01234 lose 1 game) 5 games
01234 plays 56789 (56789 lose 1 game) 5 games
01234 plays 56789 (01234 lose again, eliminating those teams) 5 games
56 plays 78 (56 lose again, eliminating those teams) 2 games
7 plays 8 (7 loses again, eliminating team 7) 1 game
8 plays 9 (8 loses again, eliminating team 8) 1 game
Leaving us with 9 as the only remaining team (with one loss)

5+5+5+2+1+1 = 19 OR
x = 10 (2*10-1) = 19
Re: Games league champion A3   [#permalink] 11 Jan 2008, 15:21
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