8 balls how to arrange for adjoining?

[Helly123]

Homework Statement

Homework Equations

The Attempt at a Solution

the answer for no adjoining
_W_W_W_W_W_
for 3 red balls, there are 6 positions
so $6C_3 = 20$
i'm curious, on other way to find arrangement?

for adjoining = all arrangement - adjoining
all arrangement = 3 red can get to any positions _ _ _ _ _ _ _ _ while white at the rest
i'm curious on other way to find arrangement?
so $8C_3 = 56$
adjoining = 56 - 20 = 36

i'm curious on how to find adjoining directly without ' all arrangement - adjoining ' ?

 

[Ray Vickson]

Homework Statement

View attachment 206649

Homework Equations

The Attempt at a Solution

the answer for no adjoining
_W_W_W_W_W_
for 3 red balls, there are 6 positions
so $6C_3 = 20$
i'm curious, on other way to find arrangement?

for adjoining = all arrangement - adjoining
all arrangement = 3 red can get to any positions _ _ _ _ _ _ _ _ while white at the rest
i'm curious on other way to find arrangement?
so $8C_3 = 56$
adjoining = 56 - 20 = 36

i'm curious on how to find adjoining directly without ' all arrangement - adjoining ' ?

Place the three red balls in a line with spaces between them; now you have 4 spaces (two between the outer and central reds and one at each end, outside). Your have 5 white balls that must be put into the 4 spaces, and two of those spaces must contain at least one white ball. The number of remaining possibilities is very limited.

 

[SammyS]

Homework Statement

View attachment 206649

Homework Equations

The Attempt at a Solution

the answer for no adjoining
_W_W_W_W_W_
for 3 red balls, there are 6 positions
so $6C_3 = 20$
i'm curious, on other way to find arrangement?

for adjoining = all arrangement - adjoining
all arrangement = 3 red can get to any positions _ _ _ _ _ _ _ _ while white at the rest
i'm curious on other way to find arrangement?
so $8C_3 = 56$
adjoining = 56 - 20 = 36

i'm curious on how to find adjoining directly without ' (all arrangement) - (non-adjoining) ' ?

Two possibilities for adjoining the red balls:

1. All three Reds are adjoining.
2. Two Reds are adjoining. The third Red is not adjoining.

For either one, use the _W_W_W_W_W_ configuration.
.

 

[Helly123]

Place the three red balls in a line with spaces between them; now you have 4 spaces (two between the outer and central reds and one at each end, outside). Your have 5 white balls that must be put into the 4 spaces, and two of those spaces must contain at least one white ball. The number of remaining possibilities is very limited.

I've seen someone do it, but I don't really get it
_R _ _ R _ R_

(why 4 spaces?)
R _ _ R _ R_

W must on at 2 of them
R_W R W R _ ( like this? )
left 3 W, what am I supposed to do?

 

[Helly123]

Two possibilities for adjoining the red balls:

1. All three Reds are adjoining.
2. Two Reds are adjoining. The third Red is not adjoining.

For either one, use the _W_W_W_W_W_ configuration.
.

yes, I've tried it
for all Reds adjoining
RRR go together at one of the place at _W_W_W_W_W_ ?
$6C_1$

for RR and R go to 2 places at _W_W_W_W_W_
$6C_2$

how am I supposed to do?

 

[Ray Vickson]

I've seen someone do it, but I don't really get it
_R _ _ R _ R_

(why 4 spaces?)
R _ _ R _ R_

W must on at 2 of them
R_W R W R _ ( like this? )
left 3 W, what am I supposed to do?

I answered only for the case of NO adjoining reds.

To answer your question: look at your first diagram, which I have re-drawn: ____R____R____R____ (making each space wide enough to hold 5 balls). How many spaces ( ____ ) do you see?

The two middle spaces must each have at least one W; that leaves 3W's to go into the 4 spaces.

Now you are supposed to figure out how many possibilities there are. Look up "combinations" on-line if your book does not do it. (You have already been told about this in the other thread about counting chords.)

I cannot offer more help without doing the whole question for you.

 

[SammyS]

...

for RR and R go to 2 places at _W_W_W_W_W_
$6C_2$

how am I supposed to do?

How many ways can you place RR ?
For each of those, how many ways can you place R ?

 

[Helly123]

I answered only for the case of NO adjoining reds.

To answer your question: look at your first diagram, which I have re-drawn: ____R____R____R____ (making each space wide enough to hold 5 balls). How many spaces ( ____ ) do you see?

The two middle spaces must each have at least one W; that leaves 3W's to go into the 4 spaces.

So i get :
If 3 W left can get to 4 positions
_R_R_R_

1st : all 3 go to same position _R_R_RWWW
So 4 arrangements

And then 2 W and 1W go to different position so we use permutation because we take the account of order, 4P2 = 12
R_RWWRW

And then 3 W go to 3 different position
WRWRW_
So 4C3 = 4

All arrangents : 4+4+12 = 20

 

[Helly123]

How many ways can you place RR ?
For each of those, how many ways can you place R ?

So 3R go to _W_W_W_W_W_

For RRR adjoining can go to 6 positions
6 arrangements

For RR and R go to different places
6P2 = 30 arrangements

So total arrangements = 36

 

[Ray Vickson]

So i get :
If 3 W left can get to 4 positions
_R_R_R_

1st : all 3 go to same position _R_R_RWWW
So 4 arrangements

And then 2 W and 1W go to different position so we use permutation because we take the account of order, 4P2 = 12
R_RWWRW

And then 3 W go to 3 different position
WRWRW_
So 4C3 = 4

All arrangents : 4+4+12 = 20

Right, or you can just quote standard results (available in lots of places on-line). You want to place $k = 3$ identical balls into $n = 4$ non-identical boxes (with no restrictions on the number of balls in any of the boxes), and the number of different ways of doing that is known to be $C_k^{n+k-1} = C^6_3 = 20$, as you already obtained.