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.