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

Thanks.

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

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!

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?

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

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

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.

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

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

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