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8 balls how to arrange for adjoining?

  1. Jul 6, 2017 #1
    1. The problem statement, all variables and given/known data

    16_Mat_A_1.6.png
    2. Relevant equations


    3. 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 ' ?
     
  2. jcsd
  3. Jul 6, 2017 #2

    Ray Vickson

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    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.
     
  4. Jul 6, 2017 #3

    SammyS

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    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.
    .
     
  5. Jul 6, 2017 #4
    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?
     
  6. Jul 6, 2017 #5
    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?
     
  7. Jul 6, 2017 #6

    Ray Vickson

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

    SammyS

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    How many ways can you place RR ?
    For each of those, how many ways can you place R ?
     
  9. Jul 7, 2017 #8
    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
     
  10. Jul 7, 2017 #9
    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
     
  11. Jul 7, 2017 #10

    Ray Vickson

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    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.
     
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