Doing a brute force count, it looks like 36...erm, I mean 243...um, I mean 120

Actually one of my early answers was 243, but that seemed like too many and I wasn't awake enough to commit to memory how I arrived at 120:

RRR--
R-RR-
R--RR
RR-R-
RR--R
R--RR
-RRR-
-RR-R
-R-RR
--RRR

that's 10 different positional combinations for any one color, meaning 30 (3*10) different positional combinations if we are given 3 and only 3 colors

for each positional combinations where 3 of one color is present, there are also 2 different positional combinations for each of the other colors

example:
GBRRR
BGRRR
RBGGG
BRGGG
RGBBB
GRBBB

that gives 60 total possibilities when 3 of one color is present

RR---
R-R--
R--R-
R---R
-RR--
-R-R-
-R--R
--RR-
--R-R
---RR

interesting...there are ALSO 10 possible positional combinations when only 2 positions are taken up by one color

That means there are 20 possible positional combinations for each combination of 2 colors, giving 60 (20*3 combinations of 2 colors[RB, RG, BG] holding 2 positions each)

That's 120 possible positional combinations when 2 or 3 colors is present

Tell me what I missed