Question

A class consists of 3 boys and 6 girls willing to form 3 groups of 3 called Groups A, B, C.
How many ways are there to assign 9 of them to Groups A, B C?
How many ways are there to assign them to the groups such that every group has one boy?
How many ways are there to assign them to groups such that all boys are in one group?
........such that only one group has no boys?

Answer 1

@jim_thompson5910

Answer 2

Is a group with 0 pupls considered valid?

Answer 3

No. All must have three.

Answer 4

Ok. I misread, sorry.
The answer for the first question is C(9,3) * C(6,3).
For the second one, you can choose 3! = 6 ways to distribute the boys and then have C(6, 2) * C(4, 2) ways to distribute the girls, therefore the answer is 3! * C(6,2) * C(4,2)
For the third one you have to choose the group that has no boys and there are 3 possibilities and afterwards you simply have to share the girls between the other 2 groups as in the first 2 examples.
For the last question, you have to choose a group to get 2 boys and a group to have one boy. Afterwards, you share the girls. It is solvable the same way as the previous ones.

Answer 5

I don't think that's right.
@campbell_st

Answer 6

@KingGeorge

Answer 7

Here's what I thought for the first one:
By partitioning: \[\frac{ 9! }{ 3!3!3! }\] but I think some other ditributing factor is missing

Answer 8

Same as \[\left(\begin{matrix}9 \\ 3\end{matrix}\right)\left(\begin{matrix}6 \\ 3\end{matrix}\right)\left(\begin{matrix}3 \\ 3\end{matrix}\right)\]

Answer 9

That's the same thing SRadu got. As far as I can tell, SRadu got the problem exactly right. What part do you think is in error?

Answer 10

Well, what about the fact that the groups have distinct names? And the students are also assumed to be distinct?
Does this affect the answer?

Answer 11

I did exactly what SRadu did, earlier, but my partners said it's wrong.

Answer 12

I don't think those should effect the answer that sradu got.

Answer 13

Someone on here said it should be \frac{ 9! }{ 3!3!3! } *3!
3! for the number of ways they can be put into groups A. B, C

Answer 14

\[ \frac{ 9! }{ 3!3!3! } *3!\]

Answer 15

That depends. The way the question is written, I don't think that's the case.

However, if you were given three groups, and you also had to name them A,B,C, then you would be multiplying by 3! on every question. The way the question is written though, makes me think you don't need to name the groups.

If possible, ask your teacher for clarification, or state which way you'll be using at the top of the problem.

Answer 16

I suppose it is possible that you are meant to apply the names to the groups as well.

In my opinion, that's the most frustrating thing about combinations/permutations. The questions can become very ambiguous.

Answer 17

My point exactly. It's really annoying because I've been working on it. The teacher asked us to get help from outside sources, but everyone seems to interprete the question dirrently.

Answer 18

So, the 3! comes in if the groups have to be labeled?
So without the 3! now, what do teh groups exist as? Floating packets of students?

Answer 19

This is the exact question:
A class consists of 3 boys and 6 girls. The studnts are randomly assigned to 3 groups of 3 labeled Group 1 Group 2 Group 3.

Answer 20

How many ways are there to assign 9 of them to Groups A, B C?
How many ways are there to assign them to the groups such that every group has one boy?
How many ways are there to assign them to groups such that all boys are in one group?

Answer 21

How many ways are there to assign them such that only one group has no boys?

Answer 22

That wording still seems to imply to me that the groups are already labeled. In which case you don't multiply by 3!.

As for what they exist as? I've never given much thought to that question :P

Answer 23

Yeah. So what'll be the solution to the last part?

Answer 24

You'll need to do a couple things for that. First, you want to choose a group to have 0 boys. There are 3 choices for that. With the two remaining groups, you want to choose a group with 1 boy. You have two options for the group, and 3 for the boy. So 6 options total, which leaves us at\[3\cdot3\cdot2\]Now we're concerned with the girls. First, choose 3 to be in the empty group. So that's \(\binom63\). And then choose 2 to be in the group with one boy. That's \(\binom32\). But then the last girl doesn't have any choice where to go, so the total is\[3\cdot3\cdot2\cdot\binom63\binom32\]Did that make sense?

Answer 25

Yes, I used a different method but got the same thing. The only problem is the naming thingy that keeps coming up.

Answer 26

Thanks a lot. I'll probably stick to what I think it is.

Answer 27

By "what I think it is", I mean this right here.

Answer 28

On the other hand, take a look at someone's solution
http://openstudy.com/users/virgil#/updates/524f3f3be4b08b5f780ca9f5

Answer 29

I would stick with this, but I can't ignore the fact that there is some ambiguousness in the way the question is worded. The way the question is stated, I'm relatively confident that the groups start with the labels 1,2, and 3. The only question then is whether you have to apply the labels A,B, and C to the groups as well.

Holy cow the wording on this is confusing.

Answer 30

Yeah. Finally, if the other method happens to be right, all I have to do is multiuply each by 3! right? Even the last part?

Answer 31

I believe so. And that's what wio was doing as well (although in a less-obvious way).

Answer 32

Yeah, took some simplification for his/her solution to decompose to mine*3!.
Well, I've tried enough. Thank you very much.

Answer 33

You're welcome.

Answer 34

@SithsAndGiggles Got any thoughts on these?