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Re: The University of Maryland, University of Vermont, and Emory [#permalink]
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MyGuruStefan wrote:
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The University of Maryland, University of Vermont, and Emory University have each 4 soccer players. If a team of 9 is to be formed with an equal number of players from each university, how many number of ways can the selections be done?

To avoid using Combination or Permutation notation, think through the problem logically.

If a team of nine is to be formed with an equal number of players from three teams, then each team must contribute 3 players. Since there are 4 players available from each team that means that in each combination only one player will be left out, which implies that at any given time one of the 4 players could be left out. Therefore, there are logically only 4 ways to select the playing players from each team.

Then, since there are 4 available combinations from each team and three teams in total, add those three outcomes together to find that there are 4 + 4 + 4 = 12 different combinations of available players, which matches choice C.


Are we looking for the number of ways to arrange the 9-player team? If so, doesn't the Fundamental Counting Principle tell us to MULTIPLY the number of ways each individual team can make the selection? In this case, I think the answer should be 4*4*4 = 64
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Re: The University of Maryland, University of Vermont, and Emory [#permalink]
64???
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Re: The University of Maryland, University of Vermont, and Emory [#permalink]
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170896 wrote:
Are we looking for the number of ways to arrange the 9-player team? If so, doesn't the Fundamental Counting Principle tell us to MULTIPLY the number of ways each individual team can make the selection? In this case, I think the answer should be 4*4*4 = 64

That would be the method for finding a THREE player team, not a nine-player team :-)
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Re: The University of Maryland, University of Vermont, and Emory [#permalink]
What will be the answer if the question like this ------

The University of Maryland, University of Vermont, and Emory University have each 4 soccer players. If a team of 9 is to be formed with an equal number of players from each university, how many possible teams are there?
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Re: The University of Maryland, University of Vermont, and Emory [#permalink]
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please let me know if i am right.
Each student is given a number, so:

Maryland University students: 1,2,3,4
Ways 3 persons possibly can be selected: (1,2,3) ,(1,2,4), (1,3,4),(2,3,4)

Vermont University students: 5,6,7,8
Ways 3 persons possibly can be selected: (5,6,7) ,(5,6,8), (5,7,8),(6,7,8)

Emory University students: 9,10,11,12
Ways 3 persons possibly can be selected: (9,10,11) ,(9,10,12), (9,11,12),(10,11,12)

Number of ways a team of 9 is formed:
(1,2,3)(5,6,7)(9,10,11)
(1,2,3)(5,6,7)(9,10,12)
(1,2,3)(5,6,7)(9,11,12)
(1,2,3)(5,6,7)(10,11,12).....
so there are 64 different teams of 9 persons selected equally from 3 Universities
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Re: The University of Maryland, University of Vermont, and Emory [#permalink]
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mohammadmh91 wrote:
please let me know if i am right.
Each student is given a number, so:

Maryland University students: 1,2,3,4
Ways 3 persons possibly can be selected: (1,2,3) ,(1,2,4), (1,3,4),(2,3,4)

Vermont University students: 5,6,7,8
Ways 3 persons possibly can be selected: (5,6,7) ,(5,6,8), (5,7,8),(6,7,8)

Emory University students: 9,10,11,12
Ways 3 persons possibly can be selected: (9,10,11) ,(9,10,12), (9,11,12),(10,11,12)

Number of ways a team of 9 is formed:
(1,2,3)(5,6,7)(9,10,11)
(1,2,3)(5,6,7)(9,10,12)
(1,2,3)(5,6,7)(9,11,12)
(1,2,3)(5,6,7)(10,11,12).....
so there are 64 different teams of 9 persons selected equally from 3 Universities


The question is not asking for the number of different teams that can be formed, in which case you are right. The question is asking how many number of ways the selections can be done.

We can select 3 out of 4 players from The University of Maryland in 4!/(3! * (4 - 3)!) = 4 ways
We can select 3 out of 4 players from The University of Vermont in 4!/(3! * (4 - 3)!) = 4 ways
We can select 3 out of 4 players from Emory University in 4!/(3! * (4 - 3)!) = 4 ways

Therefore, the total number of ways the selection can be made is 4 + 4 + 4 = 12 ways
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Re: The University of Maryland, University of Vermont, and Emory [#permalink]
Also it is not an official ETS question. The source is NOVA. Hence the confusion and ambiguity. The wording of the problem could have been better.
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Re: The University of Maryland, University of Vermont, and Emory [#permalink]
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