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 can such distinct groups be formed?

Answer 1

@hartnn @ganeshie8

Answer 2

@thomaster

Answer 3

@myininaya

Answer 4

So boys and girls are not distinct? LOL!

Answer 5

My strategy is: if you assign the girls to each group, then there is only one way to assign the remaining boys.

Answer 6

And vise versa. In fact in this case it looks like it would be easier to assign the boys into each group.

Answer 7

Distinct as in they have different names and the groups are noted. Meaning, order is important.

Answer 8

@wio Can you give a mathematical interpretation?

Answer 9

Seems like there is no individuality here. There is only distinction by gender.

Answer 10

You want to know how I'd assign the boys to the groups?

Answer 11

@wio The rest of the question is actually cut off. This is just the first part. The other parts talk about assigning boys and girls separately. For now, I just want to know how to do this part.

Answer 12

Okay let me get this straight:
The kids are distinct. The gender doesn't matter then.

Answer 13

You have \(3+6=9\) total kids.

Answer 14

Gender doesn't matter yet. Not in this part of the question. Yes, total is 9.

Answer 15

The ways to divide \(9\) kids among \(3\) groups of size \(3\) is \[
\frac{9!}{3!3!3!}
\]
http://www.wolframalpha.com/input/?i=9!%2F(3!3!3!)

Answer 16

Does that take into account that the groups have different names?
Shouldn't one multiply by 3! because of the ways they can choose names?

Answer 17

Or you could do it this way: \[
\binom 93\times \binom 63 \times \binom 33
\] http://www.wolframalpha.com/input/?i=(9+choose+3)(6+choose+3)(3+choose+3)

Answer 18

That's what I did at first. I understand how that works. I'm just concerned about the fact that they can take separate names afterward.

Answer 19

So you want it to be:
1a) choose a group
1b) choose the group members
2a) ...
?

Answer 20

Like this? \[
\binom 31\binom 93\times \binom 21\binom 63 \times \binom 11\binom 33
\] http://www.wolframalpha.com/input/?i=%283+choose+1%29%289+choose+3%29+%282+choose+1%29%286+choose+3%29+%281+choose+1%29%283+choose+3%29

Answer 21

Okay, which one is the proper way?

Answer 22

Exact question says "How many ways are there to assign 9 of them to Groups A, B C?

Answer 23

I suppose the second way then.

Answer 24

Why not just *3! ? Can you explain what you did there?

Answer 25

@wio ?

Answer 26

I chose one group before choosing it's members

Answer 27

http://www.wolframalpha.com/input/?i=%283+choose+1%29%282+choose+1%29%281+choose+1%29+-+3%21

Answer 28

How about:
Divide them into 3 groups
Assign the groups names A,B,C. *3! ?

Answer 29

You can do that too.

Answer 30

Okay, so in what case is the first methid accurate? The first one you did?

Answer 31

*method

Answer 32

Umm, the first method is accurate if the groups are indistinguishable, I believe.

Answer 33

Okay, thanks. @wio
How many ways are there to assign students to these groups such that every group includes 1 boy?

Answer 34

1a) choose a group
1b) choose a boy
1c) choose two girs

Answer 35

\[
\binom 31 \binom 31 \binom 62
\times
\binom 21 \binom 21 \binom 42
\times
\binom 11 \binom 11 \binom 22

\]

Answer 36

Are you sure? Seems quite large.
I was thinking: \[\frac{ 6! }{ 2!2!2! } *3!\]

Answer 37

Thought: Partition girls into three groups of 2. Then multiply by the number of ways the boys can fill the slots.

Answer 38

Are the groups no longer distinct?

Answer 39

They always are,

Answer 40

@Mertsj @agent0smith What do you think?

Answer 41

It's getting confusing because of the fact that the groups have names.

Answer 42

I am not good at probability but I can tell you what I think.
For the first group we have( 9 x 8 x 7)/3! which is 84
For the second group we have (6x5x4)/3! which is 20
For the last group we have (3x2x1)?3! which is 1
The grand total is 105

Answer 43

Trust my answer. =) I already explained how it works fundamentally.

Answer 44

I think wio's answers here are on-point.

Answer 45

Unless you can find where I'm double counting or something.

Answer 46

\[\large \binom 31\binom 93\times \binom 21\binom 63 \times \binom 11\binom 33\]
and \[\large \binom 31 \binom 31 \binom 62 \times \binom 21 \binom 21 \binom 42 \times \binom 11 \binom 11 \binom 22\]