Red and blue beads are being strung to make a bracelet

Visualize the beads on a string stretched out in front of you
Cases:
A) all blue --- one way

B) 1 red, 4 blue:
number of choices = 5!/(1!4!) = 5
e.g.
RBBBB, BRBBB, BBRBB, BBBRB, BBBBR
BUT... RBBBB and BBBBR are really the same bracelet, merely flipped, so there are really only 3 of these

C) 2 red, 3 blue
number of choices = 5!/(2!3!) = 10
again , 5 of those are symmetrical, so only 5 of these

D) 3red, 2 blue
same as C), --- so 5 of those

E) 4red, 1 blue
same as B), so 3 of these

F) all red ---- one of them

total = 1 + 3 + 5 + 5 + 3 + 1 = 18

check my thinking.

8. If all 5 beads are red, only 1 bracelet can be formed. If 4 beads are
red and 1 is blue, only 1 bracelet can be made. If 3 beads are red and 2 beads are blue, 2 different bracelets are possible: the blue beads together or separated by the red beads. The symmetry in this situation means that there will be 2 bracelets with 2 red and 3 blue beads, 1 bracelet with 1 red bead, and 1 bracelet with no red beads, for a total of 8 bracelets.