Rotation-distinct rings of three red, three blue

Six beads, three red and three blue, are threaded onto a circular bracelet. Two bracelets count as the same if one can be rotated into the other; reflections (turning the bracelet over) are not treated as the same, so for instance $RRBRBB$ and $RRBBRB$ are different. How many different bracelets are possible?

$4$. By Burnside's lemma, average the fixed colourings over the $6$ rotations: the identity fixes $\binom63=20$, the two rotations by a third each fix $2$, and the others fix none, so $(20+2+2)/6=4$. Concretely: all reds together, the two mirror-image '2+1' patterns, and the alternating one.