gmatusa2010 wrote: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?

Is Ans 64 Way?

Is ans 64 ways?

Here's my thought process.

Order doesn't matter, right? No matter how the group is rearranged, the same players are included once the players are selected.

Each school has to select 3 out of 4 of its players: 4!/3! = 4
There are 3 schools: 3
The group includes schools Maryland AND Vermont AND Emory: multiply
Number of Possible Groups: (4)(4)(4) = 4^3 = 64

Am I missing something?

This actually isn't a probability question; it's a combinatorics question. In order for there to be an equal number of players from each of the three teams, there needs to be three taken from each group of four. There are 4!/3!(4-3)! number of ways of taking 3 players from a group of four. This is the case because order doesn't matter, i.e. a group of three is a group of three regardless of who is in it. Since there are four ways of ordering each group of three, there are 4 x 4 x 4 = 64 ways of ordering the players. At least that's what I thought.

So we have 9 players on 3 teams and we need to divide the players evenly, therefore there should only be one case here:

Case 1: 3 players are chosen from Team A, 3 players are chosen from Team B, 3 players are chosen from Team C

Team A: Number of ways to choose 3 players from 4: 4C3 = 4

Team B: Number of ways to choose 3 players from 4: 4C3 = 4

Team C: Number of ways to choose 3 players from 4: 4C3 = 4

Number of ways we can select the 9 players: 4 * 4 * 4 = 64

I thought the same as everyone else but here's OA: (Can someone explain? Answer possibly wrong?)

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?
(A) 3
(B) 4
(C) 12
(D) 16
(E) 25
Selecting 3 of 4 players can be done in 4C3 =
4!
3!1!
= 4 ways.
The selection from the 3 universities can be done in 3 4 = 12 ways.
The answer is (C).