Place 4 green balls along the straight line.
Make gaps between these four green balls.

     You will have three gaps between the 4 green balls 
PLUS one gap on the left before these balls 
PLUS one gap on the right after these balls.


We will place 3 red balls and 2 white balls in these 3 + 2 = 5 gaps.
We MUST place some of these 3 + 2 = 5 balls in the gaps between the green balls
and we CAN place some of these 5 balls in the gaps before and/or after 4 green balls.


                It is the key idea to the problem' solution.


The rest is simply an implementation of this idea and detailed consideration/outlining/listing/counting 
of all possible cases.


    +----------------------------------------------------------------------------------------+
    |           Let the gaps between the 4 green balls are numbered 1, 2, 3;                 |
    |   the gap before green balls has number 0 and the gap after green balls has number 4.  |
    +----------------------------------------------------------------------------------------+



(1)  This is the case when all 3 red balls and 2 white balls are placed in 3 gaps between the 4 green balls.

     We have these sub-cases

               gap  #       1    2    3

        balls in the gaps   RWR  R    W,  
                            R    RWR  W,
                            R    W    RWR,   
        PLUS
                            RWR  W    R,  
                            W    RWR  R,
                            W    R    RWR,   
        PLUS
                            WRW  R    R,  
                            R    WRW  R,
                            R    R    WRW.   

     It gives  3 + 3 + 3 = 9 possible arrangements.


     We also have these other sub-cases

               gap #        1    2    3

        balls in the gaps   2    2    1,   where  '2'  is  (R,W)  or (W,R) independently at every '2' appearance.

                            2    1    2,

                            1    2    2.   It gives  = 12 possible arrangements.


    So, case (1) gives 9 + 12 = 21 different possible arrangements.



(2)  This is the case when all 3 red and 2 white balls are placed in 5 gaps, one ball in each gap.

     It gives   =  = 10 different possible arrangements for case (2).



(3)  This is the case when 2 balls of different colors are placed in the gap '0'; 
     three remaining balls of 5 = 3R + 2W balls are placed in gaps 1, 2, and 3.

     It gives these arrangements

               gap  #       0   1    2    3   4

        balls in the gaps  RW   R    R    W,
                           RW   R    W    R,
                           RW   W    R    R,      (3 arrangements)
        PLUS
                           WR   R    R    W,
                           WR   R    W    R,
                           WR   W    R    R.      (3 arrangements).


    It gives 3 + 3 = 6  different possible arrangements for case (3).



(4)  This is the case when 2 balls of different colors are placed in the gap '4'; 
     three remaining balls of 5 balls (3R + 2W) are placed in gaps 1, 2, and 3.

     This case is SYMMETRIC to case (3).  It gives 6 other arrangements, symmetric to case (3).



(5)  This is the case when 1 ball R or W is placed in the gap '0'; 
     four remaining balls of 5 = 3R + 2W balls are placed in gaps 1, 2, and 3, 
     so as two balls (R and W) are placed into the same gap.

     It gives these arrangements

               gap  #       0   1    2    3   4

        balls in the gaps   R   R    W    RW,
                            R   W    RW   R,
                            R   RW   R    W,    

                            R   W    R    RW,
                            R   R    RW   W,
                            R   RW   W    R,    

                            R   R    W    WR,
                            R   W    WR   R,
                            R   WR   R    W,    

                            R   W    R    WR,
                            R   R    WR   W,
                            R   WR   W    R      (3*4 = 12 arrangements)


        PLUS
                            W   R    R    RW,
                            W   R    RW   R,
                            W   RW   R    R,    

                            W   R    R    WR,
                            W   R    WR   R,
                            W   WR   R    R.     (3*2 = 6 arrangements)


    It gives 12 + 6 = 18  different possible arrangements for case (5).



(6)  This is the case when 1 ball R or W is placed in the gap '4'; 
     four remaining balls of 5 = 3R + 2W balls are placed in gaps 1, 2, and 3, 
     so as two balls (R and W) are placed into the same gap.

     This case is SYMMETRIC to case (5).  It gives 18 other arrangements, symmetric to case (5).



(7)  From cases (1), (2), (3), (4), (5) and (6),  we have, in all,  

         21 + 10 + 6 + 6 + 18 + 18 = 79 different possible arrangements.


ANSWER.  Doing this way, I counted 79 different possible arrangements.