Arranging $3$ red balls, $2$ blue balls, and $2$ green balls so that no two adjacent balls are of the same color I want to arrange $3$ red balls, $2$ blue balls, and $2$ green balls in a line so that no two balls of the same color are adjacent. If there are no such restrictions, then the number of such unique arrangements is 
$$\frac{7!}{3!\times 2!\times 2!}=210.$$
I found a related problem here but it does not have an answer. Instead, I wrote a code in R that will completely list all such possible arrangements. I found 38. I also tried to list them manually below. 


*

*_ B _ G _ B _ G _: $C^5_3=10$ arrangements

*_ G _ B _ G _ B _: $C^5_3=10$ arrangements

*_ G R G _ B R B _: 3 arrangements

*_ B R B _ G R G _: 3 arrangements

*_ G _ B R B _ G _: $C^4_2=6$ arrangements

*_ B _ G R G _ B _: $C^4_2=6$ arrangements


Is this correct? Is there a more elegant way of doing this?
 A: There are $10$ ways to arrange the red balls. Of these:


*

*two arrangements leave three adjacent free spots and one solo free spot:
$$
R\,\_\,\_\,\_R\,\_R\\
R\,\_R\,\_\,\_\,\_R
$$For each of these: Whichever colour is in the middle of the three free spots must also be in the solo spot, so 2 possibilities for each of these arrangements

*six arrangements leave two adjacent spots and two solo spots:
$$
\,\_\,\_R\,\_R\,\_R\\
\,\_R\,\_\,\_R\,\_R\\
\,\_R\,\_R\,\_\,\_R\\
R\,\_\,\_R\,\_R\,\_\\
R\,\_R\,\_\,\_R\,\_\\
R\,\_R\,\_R\,\_\,\_
$$ 
and one leaves two pairs of adjacent spots:
$$
R\,\_\,\_R\,\_\,\_R
$$For each of these: the colours must differ in a pair of adjacent spots, but they can be swapped around, so 4 possibilities for each of these arrangements

*one arrangement leaves no adjacent spots: 
$$
\,\_R\,\_R\,\_R\,\_
$$
Here there are six possible arrangements of the remaining $4$ balls
In total:
$$
2\cdot 2 + 7\cdot 4 + 1\cdot 6 = 38
$$
I don't think there is a more "elegant" way to do this than simply splitting into cases and count. The interactions are too complicated to make inclusion-exclusion viable, and there isn't a simple formula for "non-adjacent" that we can apply. Of course, how you decide to split into cases changes how easy the calculations are and how easily you miss a case or double count.
A: You might find applying inclusion-exclusion more "elegant"
Keeping the A's apart throughout, count the # of arrangements with
(A's apart) - (A's apart, B's or C's bunched)   + (A's apart, both B's and C's bunched)
$ = \left[\frac{4!}{2!2!}\times \binom53 \right] - \left[2\times \frac{3!}{2!}\binom43\right] + \left[2!\times\binom33\right] = 60 - 2\cdot12 + 2 =38$ 
