A certain league has four divisions. The respective divisions had 9, 1

Question

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

(A) 79
(B) 83
(C) 85
(D) 87
(E) 88

KarishmaB · Accepted Answer

Actually solving the problem doesn't take very long. Think of it this way:

We need to keep track of losses. Let's focus on those and forget about the wins. 
Every time a game is played, someone loses. You can give at most 2 losses to a team since after that it is out of the tournament. 
Consider the division which has 9 teams. What happens when 18 games are played? There are 18 losses and each team gets 2 losses (you cant give more than 2 to a team since it gets kicked out after 2 losses) so all are out of the tournament. But we need a winner so we play only 17 games so that the winning team get only 1 loss.

Similarly, the division with 10 teams can have at most 2*10 - 1 = 19 games.
The division with 11 teams can have at most 2*11 - 1 = 21 games.
The division with 12 teams can have at most 2*12 - 1 = 23 games.
This totals up to 80 games (note that the average of 17, 19, 21 and 23 will be 20 so the sum will be 4*20 = 80).

Now you have 4 teams. 1 loss gets a team kicked out. If you have 3 games, there are 3 losses and 3 teams are kicked out. You have a final winner!
Hence the total number of games = 80 + 3 = 83

stolyar · Answer

A hard nut...

As for divisional games: a minus is given for losing a game.

the 9-team division) lets count minuses; 8 teams can have 2 minuses and one (a champion) can have 1 minus. Total 17 minuses.

the 10-team division) lets count minuses; 9 teams can have 2 minuses and one (a champion) can have 1 minus. Total 19 minuses.

the 11-team division) lets count minuses; 10 teams can have 2 minuses and one (a champion) can have 1 minus. Total 21 minuses.

the 12-team division) lets count minuses; 11 teams can have 2 minuses and one (a champion) can have 1 minus. Total 23 minuses.

After that, four teams remained. I assume that when 3 teams can have 1 minus and the overall champion has none. Total 3 minuses.

Overall, there can be 83 minuses. Tus, it is B.

manpreetsingh86 · Answer

Let's name the teams in group 1 as 1,2,3,4,5,6,7,8,9.

Case 1; team1 played with every other team and won all of its matches.
so total number of matchs =8
case 2: team2 , played with team 3,4,5,6,7,8,9 and won all of its matches.
total number of matches =7
 after case 1 and case 2 we have only two teams remaining in the group1 which are team 1 and team 2. Now since question asks us for the maximum no. of matches. Therefore we must include the extra case in which team 2 defeated team 1. Now both team 2 and team 1 have 1 loss each. Now in the final match, we will found out about the eventual winner in group 1.
 maximum no. of matches in group 1 are 8+7+1(in which team2 defeated team1) + 1 ( final) =17

Similarly in group 2 we have 9 + 8 +1 +1 =19
group 3 = 10+9+1+1 =21
group 4 = 11+10+1+1=23

After this we will have 4 winner from each group. lets name them as w1,w2,w3,w4
Let's assume w1 won all of its matches from the remaining three teams and eventually emerged as a winner. Therefore total matches among four winners=3

Therefore maximum total no. of matches played are 17+19+21+23+3=83

ShantnuMathuria · Answer

Dear Manpreet

I was applying the same trick but got confused in between.

Since, we allowed 3 matches between team 1 and team 2 in the first group, are you assuming that more than one match can be played between two teams as long as the team does not lose 2 matches in total?

Thanks & regards

manpreetsingh86 · Answer

yes, you are right. see the question asks us to find  the maximum number of games that could have been played  to determine the overall league champion. hence we go for the additional matches.

i hope it is clear now.

SVaidyaraman · Answer

The intuition is each team is eliminated after 2 losses. It has to lose 2 games and it can lose only 2 games. The winner loses only once or may not lose at all. But since maximum number of games is asked we will assume that the winner loses once.

Each loss corresponds to 1 game played.

For the 4 divisions the number of games played will be 2*8 + 1, 2*9 +1 , 2*10 +1 and 2*11+1= 17,19,21,23 resp= 80 total

Applying the same logic to the finals, the maximum number of games played will be 1*3 + 0=3

Total = 80+3=83

SachinWordsmith · Answer

Hi,

Could someone comment on how they would solve/approach it if the question were : What's the minimum number of games that could be played before a winner is determined?

KarishmaB · Answer

Use the same logic as that used for maximum number of games. You know that each team must lose in at least 2 games (and at most 2 games). The only difference will be that the qualifying team will not lose any match to minimize the number of matches.

Now can you come up with the answer?

SachinWordsmith · Answer

Thanks for the Reply Karishma. I follow your posts and your explanations are top notch.

If I went with the same logic, the answer would be 4 lesser (eliminating the number of games by the group winners lost in the Prelims) = 79 Games. Can I generalize and use the above presented logic for say, a single elimination ( 1 Loss only) or triple elimination (3 Losses) format too?

KarishmaB · Answer

Yes, you are correct. 4 fewer games is all you can afford.

Whether you can generalize will depend on the question. If the question remains the same:
In case of a single loss elimination, there will be fixed number of games which will be played so there will be no maximum - minimum. Every game will have a loss and will eliminate exactly one team. 
In case of 3 loss elimination format, the qualifying team will not suffer any losses if we want to minimize the number of games and it will suffer 2 losses in case we want to maximize the number of games.

rmselva · Answer

Hi,

I had a crazy thought.
Here our goal is to maximize the matches.
In Divison A we have 9 teams
let them be a,b,c,d,e,f,g,h,i
I let team "a" play with all other teams twice
and lose every time. No. of matches=16
Similarly for b=14,c=12 etc.
So we get a total of 72 matches in Group A alone.

KarishmaB · Answer

Here is what the question says: 
"Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games --"

When a loses its matches against b and c, it will be eliminated. It will not play any other game. It doesn't need to lose 2 matches against the same team for every team.

KarishmaB · Answer

Responding to a pm:

Note that every time a team wins, another team loses. So if one of the 9 teams wins 2 matches, someone loses those 2 matches. So 2 losses of the rest of the 8 teams are already accounted for. Say one team loses those two matches. So there can be only 7 teams losing 2 matches each so that we have 14 more losses which gives us a total of 16 games played. 
Instead, let's make the winning team lose a match too. It won't be kicked out but another match would be played.

anairamitch1804 · Answer

There are two different approaches to solving this problem. The first employs a purely algebraic approach, as follows: Let us assume there are n teams in a double-elimination tournament. In order to crown a champion, n – 1 teams must be eliminated, each losing exactly two games. Thus, the minimum number of games played in order to eliminate all but one of the teams is 2(n – 1). At the time when the (n – 1)th team is eliminated, the surviving team (the division champion) either has one loss or no losses, adding at most one more game to the total played. Thus, the maximum number of games that can be played in an n-team double-elimination tournament is 2(n – 1) + 1. There were four divisions with 9, 10, 11, and 12 teams each. The maximum number of games that could have been played in order to determine the four division champions was (2(8) + 1) + (2(9) + 1) + (2(10) + 1) + (2(11) + 1) = 17 + 19 + 21 + 23 = 80.

The 9-team division) lets count minuses; 8 teams can have 2 minuses and one (a champion) can have 1 minus. Total 17 minuses((2(8) + 1)).

The 10-team division) lets count minuses; 9 teams can have 2 minuses and one (a champion) can have 1 minus. Total 19 minuses((2(9) + 1)).

the 11-team division) lets count minuses; 10 teams can have 2 minuses and one (a champion) can have 1 minus. Total 21 minuses((2(10) + 1)).

the 12-team division) lets count minuses; 11 teams can have 2 minuses and one (a champion) can have 1 minus. Total 23 minuses((2(11) + 1)).

Now you have 4 teams. 1 loss gets a team kicked out. If you have 3 games, there are 3 losses and 3 teams are kicked out. You have a final winner!
Hence the total number of games = 80 + 3 = 83.

ShashankDave · Answer

Hi  VeritasPrepKarishma

Looks very easy after one sees the solution. I actually got lost on making sequences for players in group 1(9 players), such that any player loses the second match as late as possible, so that he plays more games before getting eliminated. In this way, I got completely lost in this question. Please help me understand these types of questions. Please suggest somewhere to read so I understand every bit of these types of problems.

KarishmaB · Answer

What you have to remember is that when you are trying to make a team win, the other team is losing. So somewhere the losses are getting stacked up and those teams are getting eliminated. To you, all the teams are the same. So you should focus only on the number of losses. Each loss represents one game.

I would suggest you to not worry too much about this question. It is a tricky 700 level combinations problem. You would see at most one such question and you will need to apply a different logic to each. That is the thing about 700 level questions - you need to "figure out" each one at run time. Your best bet is to practice as many high level questions as possible to make this more intuitive.

JeffTargetTestPrep · Answer

Let’s call the four divisions A, B, C, and D, with 9, 10, 11, and 12 teams, respectively. Now let’s analyze the maximum number of games that can be played in each division.

Let’s take division A, for example. It has 9 teams and only 1 is the winning team. So, there must be 8 losing teams, and each of these teams must lose twice since it’s a double-elimination tournament. The winning team can still lose once (but not twice; otherwise it’s out of the tournament). Therefore, the maximum number of games played in division A is 8 x 2 + 1 = 17.

Notice that 8 = 9 - 1; thus, using the same argument, the maximum number of games played in each remaining division is:

Division B: (10 - 1) x 2 + 1 = 18 + 1 = 19

Division C: (11 - 1) x 2 + 1 = 20 + 1 = 21

Division D: (12 - 1) x 2 + 1 = 22 + 1 = 23

Thus, the maximum total number of games played in the 4 divisions before the single-elimination is 17 + 19 + 21 + 23 = 80.

In the single-elimination, only 3 games will be played since there will be 2 semi-finals and 1 final. (For example, in the semi-final 1, A’s champion vs. B’s champion; in the semi-final 2, C’s champion vs. D’s champion; and in the final, the winner between A and B takes on the winner between C and D to determine the overall league champion.)

Thus, the maximum total number of games played to determine the overall league champion is 80 + 3 = 83.

Answer: B

EMPOWERgmatRichC · Answer

Hi All,

While this question is 'wordy', it's a relatively straight-forward 'concept' question that doesn't require any difficult math to solve.

In a double-elimination tournament, a team that loses 2 times is eliminated, so every team EXCEPT for the 'champion' will lose twice (and the champion will lose either 0 or 1 times). We're asked to MAXIMIZE the number of games played, so we'll need each champion to lose 1 time.

Double-elimination Division games: 
9 teams = (8 teams)(2 losses each) + 1 loss for the champ = 17 games 
10 teams = (9 teams)(2 losses each) + 1 loss for the champ = 19 games 
11 teams = (10 teams)(2 losses each) + 1 loss for the champ = 21 games 
12 teams = (11 teams)(2 losses each) + 1 loss for the champ = 23 games

In the single-elimination tournament for the 4 champions, 3 of the teams will lose once and the winner will lose 0 times. That will require 3 more games total.

Maximum total games: 17 + 19 + 21 + 23 + 3 = 83 total games

Final Answer:  B

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

Vubar · Answer

I see how this is a counting problem, but I'm a little confused about where the "combinations" comes in? Is there a solution that uses the actual combinatorics theories? As it stands, we seem to be doing simple algebra, rather than using any permutations or combinations?

A certain league has four divisions. The respective divisions had 9, 1

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A certain league has four divisions. The respective divisions had 9, 1

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