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A class is divided into four groups of four students each.

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A class is divided into four groups of four students each.  [#permalink]

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New post 02 Feb 2012, 11:07
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A class is divided into four groups of four students each. If a project is to be assigned to a team of three students, none of which can be from the same group, what is the greatest number of distinct teams to which the project could be assigned?

(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)

My solution:

4C3 * 4 = 4^4 .
Is the formula used rightly selected although the answer agrees to OA?
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Re: Selecting a team from four groups  [#permalink]

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New post 02 Feb 2012, 11:22
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docabuzar wrote:
Q. A class is divided into four groups of four students each. If a project is to be assigned to a team of three students, none of which can be from the same group, what is the greatest number of distinct teams to which the project could be assigned?

(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)

My solution:

4C3 * 4 = 4^4 .
Is the formula used rightly selected although the answer agrees to OA?


Basically we are asked to determine the # of different teams of 3 that can be formed so that no 2 members are from the same group.

\(C^3_4=4\) - choosing which 3 groups out of 4 will provide with a member;
\(C^1_4*C^1_4*C^1_4=4^3\) - choosing each member from the selected 3 groups ;

So, total # of different groups will be: \(4*4^3=4^4\).

Answer: B.

P.S. Answer is correct, formula and math is not: \(C^3_4*4=4^2\).
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Re: A class is divided into four groups of four students each.  [#permalink]

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New post 04 Sep 2016, 22:28
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The problem can be solved in relatively simpler way without using combinations formula but by simply using counting method.

We have to fill 3 places by selecting 3 students from 4 groups of 4 such that no student is from same group.
Lets create groups:
1st group - a,b,c,d
2nd group - e,f,g,h
3rd group - i,j,k,l
4th group - m,n,o,p
The first position can be filled in 16 ways (a,b....p), 2nd can be filled in 12 ways - since 2nd student should not be from the same group (e,f....p), 3rd position can be filled in 8 ways (i,j,....p)
So number of ways they can be arranged is 16*12*8 = 4*4*4*3*4*2
This gives the number of arrangements where the order is important, but for us order is not important - It does not matter to us whether we select 'a' first of 'g' second... All that is important to us is that we have to select 3 people.
So we have to un-arrange the number of arrangements. We can do so by dividing the total number of arrangements with 3!.
So the final answer is 4*4*4*3*4*2/3*2 = 4^4.

Hence (B).
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Re: A class is divided into four groups of four students each.  [#permalink]

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New post 02 Feb 2012, 11:34
Thanks.

Its clear now. I m making the silly mistake of reading my 4*4 as 4^4 from the OA.
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Re: A class is divided into four groups of four students each.  [#permalink]

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New post 11 May 2017, 06:45
Can I do it the other way around: All permissible - non permissible
All permissible = 16C3 = 560
Not permissible = (4C2*14C1 - 4C3*4) by subtracting 4C3*4, I aimed to elimination double-counting
Answer I got is 240, which is wrong. But I don't know what went wrong in this logic. Pls advise, thanks!
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Re: A class is divided into four groups of four students each.  [#permalink]

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New post 11 May 2017, 06:48
another way to do it is to think of it this way:
there are 16 choices to choose from, after 1 person is chosen, we are left with 12 choices (cannot choose anyone from the same team), after that we are left with 8 choices. So 16*12*8 but then we have to divide this by 3! to elimination duplication.
So the answer is 16*16 = 4^4. Is that correct?
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Re: A class is divided into four groups of four students each.  [#permalink]

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New post 16 May 2017, 17:14
docabuzar wrote:
A class is divided into four groups of four students each. If a project is to be assigned to a team of three students, none of which can be from the same group, what is the greatest number of distinct teams to which the project could be assigned?

(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)


Each group will contribute no more than one team member. Since none of the students in the team can be from the same group, we first choose three groups from the four available groups. This can be done in 4C3 = 4!/(3!x1!) = 4 ways.

After choosing three groups, we must choose one student from each group, which can be done in 4C1 = 4 ways.

Since there are four choices for the three groups to be selected and four choices for a person to be selected in each of the three groups, in total, there are 4 x (4 x 4 x 4) = 4^4 choices.

Answer: B
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Re: A class is divided into four groups of four students each.  [#permalink]

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New post 06 Jan 2019, 12:43
Hi Bunuel,

Hi encountered the similar question in e-gmat's test. I solved the question in a slight different manner. I am not sure why my approach is incorrect. Could you please check and comment.

My approach :

I have to select three students. First student can be selected in 16 ways, second in 12 ways and then third in 8 ways. Could you please tell where am I going wrong?
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Re: A class is divided into four groups of four students each. &nbs [#permalink] 06 Jan 2019, 12:43
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