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

Question

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)

(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?

Alok322 · Accepted Answer

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).

Bunuel · Answer

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 .

docabuzar · Answer

Thanks.

Its clear now. I m making the silly mistake of reading my 4*4 as 4^4 from the OA.

kerin · Answer

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!

kerin · Answer

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?

JeffTargetTestPrep · Answer

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

dabaobao · Answer

Official Solution  
Credit: Veritas Prep

We need to make a team here. There is no arrangement involved so it is a combinations problem. First we will select 3 groups and then we will select one student from each of those 3 groups.

In how many ways can we select 3 groups out of a total of 4? From the theory discussed above, I hope you agree that we can select 3 groups out of 4 in 4∗3∗2/3!=4 ways. The interesting thing to note here is that selecting 3 groups out of 4 is the same as selecting 1 group out of 4. Why? Because we can think of making the selection in two ways – we can select 3 groups from which we will pick a student each or we can select 1 group from which we will not select a student. This will automatically give us a selection of 3 groups. We know that we can select 1 out of 4 in 4 ways (hence the calculation done above was actually not needed).

Now from each of the 3 selected groups, we have to pick one student. In how many ways can we select one student out of 4? In 4 ways. This is true for each of the three groups. We can select 3 groups and one student from each one of the three groups in 4∗4∗4∗4=4^4 ways.

ANSWER: B

CEdward · Answer

Best way to think about this is sequentially, don't panic.

1. Since we can only choose 3 students (each of which has to be from a separate group), then that means 3 groups must be chosen. 
4C3 = 4 <--- # of ways to choose 3 groups from 4

2. Within each group (of 4) we must select one
4C3 = 4

3. Total # of ways we can generate a team of 3 from 4 groups of 4:

4 x 4 x 4 x 4 = 64 = 4^4

Answer is B.

Geet03 · Answer

The selection is a 2 step one:

1st step is to select 3 groups out of 4 -> 4c3 ways
2nd step is to select one student from a group of 4 (thrice i.e., once for each group) -> 4C1*4C1*4C1

this is an "AND"scenario hence multiply :
4C3*4C1*4C1*4C1 = 4*4*4*4 = 4^4

Sans8 · Answer

I got the answer,
But if I do it something like:
 16C3-4C1*4C3

16C3--> No. of ways of selecting 3 students from 16.
4C1*4C3--> No. of ways of selecting 1 group out of 4 and then all three students from that group.
Not getting answer through this approach.
Where am I wrong?

Bunuel   bb   chetan2u

chetan2u · Answer

Hi

You are missing out on lot other possibilities. 
I will not say this is the most efficient way to do the question.

But if you want to do this, the way would be
1) Total ways to select - 16C3 or 560
2) Ways all three are from same group-4C1*4C3 or 16
3) Ways two are from the same group - (4C1*4C2)*(3C1*4C1)=288
Answer = 560-16-288=256

Direct method. 
Select any three in 4C3 ways and select one from each in 4C1 ways = 4C3*4C1*4C1*4C1 = 256

pierjoejoe · Answer

very cool problem. here we have to consider the groups like individuals to choose from.

16 students, 4 groups (G1, G2, G3, G4) of 4 students each
we want to form N TEAMS such that each TEAM member is from a different group
we have to consider first of all how to pick 3 groups out of 4 without repetition
(G1, G2, G3, __)
(G1, G2, __, G4)
(G1, __, G3, G4)
(__, G2, G3, G4)
--> because we want than to pick from each of these one student to form a TEAM.
we can pick 3 groups out of 4 in 4C3 = 4 ways (as illustrated above)

let's say we picked the groups G1, G2 and G3 (the first arrangement)
each time we pick a group we imagine we are also randomly picking a student. how many ways we can pick a student in a group? in 4 possible ways. so given that we have to pick the first group (G1)(we can pick 4 students), the second group (G2)(so we can pick 4 students) and the third group (G3) (so we can pick 4 students) --> we have a total of 4^3 possible ways to pick a 3 students from 3 different group to form a TEAM (for the counting principle)

but we have 4 possible ways to pick 3 groups from 4 --> so we make the same reasoning for the other 3 possible arrangements--> we will have at the end 
4*4^3 = 4^4 possible ways to create a TEAM that contains students from different groups.

VaibhavKulkarni · Answer

The simplest way to solve this would be,

For member 1: In how many ways can I choose 1st from 16 people = 16C1
For member 2: In how many ways can I choose 2nd from the remaining 12 people (as we can't choose from the same grp.) = 12C1
For member 3: In how many ways can I choose 3rd from the remaining 8 people (as we can't choose from the same grps) = 8C1

Therefore, (16C1*12C1*8C1)/3!

> 3! because the order of the group doesn't matter.

=> (16C1*12C1*8C1)/3! = (16*12*8)/3! = (4*4)*(4*3)*(4*2)/(3*2) =  4^4

Hope this helps!

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

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