The number of distinct bracelets of five beads made up of red, blue, and green…

Concept

Burnside's orbit-counting lemma: when a finite group G acts on a finite set X, two arrangements that some element of G carries onto each other represent the same object. The number of distinct objects equals the average number of arrangements left unchanged by a group element — that is, (1/|G|) times the sum of |Fix(g)| over every g in G, where Fix(g) is the set of arrangements that g leaves untouched.

Dividing |X| by |G| directly is valid only when every orbit has the full size |G|. Arrangements that a non-identity symmetry leaves unchanged sit in smaller orbits, so they must be counted separately — which is exactly what the lemma does.

Application to this bracelet

X is the set of colourings of five positions placed around a circle, each position independently red, blue or green, so |X| = 35 = 243. The stem treats two bracelets as the same exactly when a rotation carries one onto the other, so G is the cyclic group of the five rotations of the circle and |G| = 5.

1. Rotation by 0 positions (the identity) moves nothing, so it leaves every colouring unchanged: |Fix| = 35 = 243.

2. Rotation by 1, 2, 3 or 4 positions: because 5 is prime, each of these permutes the five positions in one single 5-cycle, so a colouring it leaves unchanged must give all five beads one and the same colour: |Fix| = 3 for each of these four rotations.

3. Sum over the whole group: 243 + 4 × 3 = 243 + 12 = 255.

4. Average over the group: 255 ÷ 5 = 51, so the bracelet count is 51.

Cross-check by direct classification

The 3 monochromatic colourings are left unchanged by every rotation, so each one forms an orbit of size 1. The remaining 243 − 3 = 240 colourings have no rotational symmetry at all, so they fall into orbits of exactly 5, giving 240 ÷ 5 = 48 classes. Altogether 48 + 3 = 51, which matches the lemma.

Why plain division fails, and one caution

Dividing straight away gives 35 ÷ 5 = 48.6, not a whole number — which already shows that plain division by the group order is invalid, because not every orbit has the full size 5. The 5-cycle analysis above is what identifies exactly which colourings sit in those smaller orbits: the monochromatic ones. Note also that if turning the bracelet over (a reflection) were allowed as well, the acting group would be the dihedral group of order 10 and the count would be (255 + 5 × 33) ÷ 10 = 390 ÷ 10 = 39. The stem restricts the identification to rotations alone, so the count here is 51.