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31 May 2022, 06:00
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
26% (02:16) correct
74%
(02:11)
wrong
based on 53
sessions
History
Date
Time
Result
Not Attempted Yet
A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won 10 games and lost 10 games; there were no ties. How many sets of three teams {A, B, C} were there in which A beat B, B beat C, and C beat A?
(A) 385
(B) 665
(C) 945
(D) 1140
(E) 1330
Are You Up For the Challenge: 700 Level Questions
(A) 385
(B) 665
(C) 945
(D) 1140
(E) 1330
Are You Up For the Challenge: 700 Level Questions
31 May 2022, 15:11
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Bunuel
Breaking Down the Info:
First of all, there are 21 teams since each team has to play each other in a single round-robin. If each team played 20 games, then there must be 21 teams in total.
Then there must be 21C2 = 21*20/2 = 210 games played in total.
Assume each team played 40 times so that each team won 20 games and lost 20 games. We can observe that each team must have won every team once and lost to every team once.
Therefore if we pick any three teams, we will have one and only one set of games such that A beat B, B beat C, and C beat A. Thus we only need to pick three teams out of the 21 for the number of combinations. (Also note that set notation does not care about order).
Thus 21C3 = 21*20*19/3! = 7*10*19 = 1330. However, this is if each team played 40 games. The reality is they only played 20 games, thus we cut this number in half (665) to find the real answer.
Answer: B
01 Jun 2022, 04:45
Kudos
Bookmarks
Every team played 20 times. Since they can't play against themselves, there must be 21 teams.
Groups of 3 teams can be selected
21!/3!18! = 1330 ways
Each will have played the other during the course of the round robin.
For example, AB, BC, CA.
There are 2 general outcomes.
One player can beat each of the other two. Those two players facing off will create another winner and loser. So one team will win two, one lose two and one winning and losing one game.
Or, each team can have one win and one loss each. How many of those groups of 3 exist is the question.
In the first instance described above, one team wins twice. So, that means 2 of that team's winning games have been selected to be in the group. This can be done
10!/2!8! = 45 ways multiplied
by 21 players =
945 ways
So, subtracting 945 from 1330 = 385 ways must be the number of ways left over to describe the scenario in the question
Trying to find the answer directly rather than by subtraction seems impossible without laying out all the potential outcomes
Posted from my mobile device
Groups of 3 teams can be selected
21!/3!18! = 1330 ways
Each will have played the other during the course of the round robin.
For example, AB, BC, CA.
There are 2 general outcomes.
One player can beat each of the other two. Those two players facing off will create another winner and loser. So one team will win two, one lose two and one winning and losing one game.
Or, each team can have one win and one loss each. How many of those groups of 3 exist is the question.
In the first instance described above, one team wins twice. So, that means 2 of that team's winning games have been selected to be in the group. This can be done
10!/2!8! = 45 ways multiplied
by 21 players =
945 ways
So, subtracting 945 from 1330 = 385 ways must be the number of ways left over to describe the scenario in the question
Trying to find the answer directly rather than by subtraction seems impossible without laying out all the potential outcomes
Posted from my mobile device
