# 8 balls how to arrange for adjoining?

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1. Jul 6, 2017

### Helly123

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
_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?

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 ' ?

2. Jul 6, 2017

### Ray Vickson

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.

3. Jul 6, 2017

### SammyS

Staff Emeritus
Two possibilities for adjoining the red balls:
1. All three Reds are adjoining.
For either one, use the _W_W_W_W_W_ configuration.
.

4. Jul 6, 2017

### Helly123

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?

5. Jul 6, 2017

### Helly123

yes, I've tried it
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?

6. Jul 6, 2017

### Ray Vickson

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.

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

Last edited: Jul 6, 2017
7. Jul 6, 2017

### SammyS

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

8. Jul 7, 2017

### Helly123

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

9. Jul 7, 2017

### Helly123

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

10. Jul 7, 2017

### Ray Vickson

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.