Question

A class consists of 3 boys and 6 girls willing to form 3 groups of 3 called Group A, B, C.
How many ways are there to create groups such that every group includes 1 boy?
How many ways are there to create groups such that all three boys are in the same group?
How many ways are there to form groups such that one group has no boys?

Answer 1

@Zarkon

Answer 2

show me what you have so far and I'll tell you if it is correct

Answer 3

For the first part, my idea is to count the number of ways to partition girls into 3 group of 2, then multiply by 3! for the number of ways the boys can fill the spaces. I'm not sure.

Answer 4

\[\frac{ 6! }{ 2!2!2! } * 3!\]

Answer 5

@Zarkon

Answer 6

@hartnn

Answer 7

I think you might be double counting

Answer 8

Can you show where?

Answer 9

(lets simplify the problem just a little) how many ways would you have if you were making 3 groups of 2...having 3 boys and 3 girls (again each group containing 1boy)

Answer 10

Meaning each group contains 1 boy and 1 girl.

Answer 11

yes

Answer 12

Number of ways of choosing 3 girls into 3 groups = 3!
Multiples by the number of ways of choosing 3 boys into 3 groups = 3!
(3!)*(3!)

Answer 13

b1g1,b2g2,b3g3
b1g1,b2g3,b3g2
b1g2,b2g1,b3g3
b1g2,b2g3,b3g2
b1g3,b2g1,b3g2
b1g3,b2g2,b3g1


did i miss any

Answer 14

b1g2,b2g3,b3g2 should be b1g2,b2g3,b3g1

Answer 15

you are saying then that there are 3!*3!=36...which are we missing?

Answer 16

oh...if you are naming the groups ...then you are correct...if the groups are indistinguishable then you would need a different answer

Answer 17

Yes the groups are named.

Answer 18

then your answer to the first one is correct

Answer 19

Are you very sure?

Answer 20

yes

Answer 21

The second part is where it all falls apart in my brain.

Answer 22

Thank you though.

Answer 23

Here's my idea:

Answer 24

if all three boys are in the same group, find the ways of putting the girls into 2 groups of 3, then find the number of ways of putting the boys as a whole into the last group.

Answer 25

@Zarkon

Answer 26

the answer will look a lot like your first answer

Answer 27

Can I get a hint?

Answer 28

I would first determine how many ways to get the girls into the first 2 groups and the boys in the last group

Answer 29

\[\frac{ 6! }{ 3!*3! } * Some factor for boys\]

Answer 30

there is only one way to select the three boys for a group of 3

\[{3\choose 3}\]

Answer 31

Yes, I got that fact, but the group of boys can take 3 different names. So *3?

Answer 32

yes

Answer 33

Is it 3 or 3!?

Answer 34

with 3! you would be double counting (because of the girls)

Answer 35

Even if the groups have names?

Answer 36

when in doubt reduce the problem so something simple


what if you had 2 girls and 1 boy

and you made 3 groups of 1

then

A B C

g1,g2,b
g2,g1,b
g1,b,g2
g2,b,g1
b,g1,g2
b,g2,g1

the number of ways is

\[\frac{2!}{1!1!}\times 3=6\]

Answer 37

Okay, thanks.
Now the last part.

Answer 38

To have exactly 1 with no boys, some group must have 2 boys while another must not have 3 boys.

Answer 39

if it is exactly one boy then

one group will have none, one group will have 1 and one group will have 2

Answer 40

Yes, but I want to do it using the result from the previos part so it'll be quicker.

Answer 41

So, my idea is: n(1 group with none) - n(One group with 3)

Answer 42

total-n(1 group with 3 boys)-N(each group with exactly 1)

Answer 43

Yeah, now I see where I flunked. I knew there was a relationship.
Are you very sure though?

Answer 44

I did it both ways...direct and with the 'complement' and I get the same answer

Answer 45

And what is total?
Can you share your answers?

Answer 46

Look at

\[\frac{6!}{1!3!2!}\cdot\frac{3!}{2!1!}\cdot 3!\]

and

\[\frac{9!}{3!3!3!}-\frac{6!}{3!3!}\cdot 3-\frac{6!}{2!2!2!}\cdot 3!\]

Answer 47

can you parse what I wrote?

Answer 48

do you understand?

Answer 49

Yes.

Answer 50

Wait, define the first part.

Answer 51

the \(\dfrac{6!}{1!3!2!}\) is the number of ways to arrange the girls

the \(\dfrac{3!}{2!1!}\) is the number of ways to arrange the boys

and the \(3!\) is the number of ways to arrange these into the 3 labeled groups.

Answer 52

ok?

Answer 53

Yes, I'm just verifying the answer. Give me a minute please.

Answer 54

I'm not getting the same answer.

Answer 55

you should

Answer 56

Total is 9!/(3!)^3 right?

Answer 57

yes

Answer 58

Check to make sure both are the same. One aboyt twice the other.

Answer 59

they are both 1080

Answer 60

jas

Answer 61

they are both 1080...

type

9!/(3!^3)-(6!/((3!)^2)*3+6!/(2!)^3*3!)

and

6!/(1!*3!*2!)*3!/(2!*1!)*3!

into wolfram alpha

Answer 62

or just copy and paste

Answer 63

I have to go...I'll be back later to see how things went.

Answer 64

Okay, thanks a lot.

Answer 65

@Zarkon I have a question about your solution.