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The University of Maryland, University of Vermont, and Emory
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14 May 2019, 00:39
14 May 2019, 00:39
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The University of Maryland, University of Vermont, and Emory University hav e 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?
(A) 3
(B) 4
(C) 12
(D) 16
(E) 25
(A) 3
(B) 4
(C) 12
(D) 16
(E) 25
GRE Prep Club Tests Editor
Joined: 13 May 2019
Affiliations: Partner at MyGuru LLC.
Posts: 186
Given Kudos: 5
Location: United States
GMAT 1: 770 Q51 V44
GRE 1: Q169 V168
WE:Education (Education)
Re: The University of Maryland, University of Vermont, and Emory
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15 May 2019, 05:43
15 May 2019, 05:43
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Expert Reply
Quote:
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.
Re: The University of Maryland, University of Vermont, and Emory
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22 Aug 2019, 18:21
22 Aug 2019, 18:21
1
Carcass wrote:
The University of Maryland, University of Vermont, and Emory University hav e 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?
(A) 3
(B) 4
(C) 12
(D) 16
(E) 25
(A) 3
(B) 4
(C) 12
(D) 16
(E) 25
We have 4 people and we're choosing 3. We're doing that 3 times.
4c3*3 = 12
Answer is C.
