P n C tough one

[rishianand7]

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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Hi rishianand7,

Many people would attempt this question using really complex math steps, but you don't have to treat it as a Perm/Comb question. Instead, just think logically:

In a 9 team league, there can be only one champion; that means that 8 teams have to lose (2 games apiece in this question)
8 x 2 = 16 games to knock out those 8 teams. HOWEVER, we want the maximum number of games, so the league champion COULD HAVE LOST 1 GAME and still been champion.

9 teams = 16 + 1 = 17 games maximum

You can repeat this process for 10, 11 and 12 team leagues:

10 teams = 18 + 1 = 19 games maximum

11 teams = 20 + 1 = 21 games maximum

12 teams = 22 + 1 = 23 games maximum

So far, that's a total of 80 games....

Now we have to factor in a single elimination set of games among the 4 champions. Since there can be only 1 champion, each of the other teams has to lose. That would take 3 games to knock out the 3 losing teams.

80 + 3 = 83 games maximum

Final Answer: B

GMAT assassins aren't born, they're made,
Rich