NECKLACE PROBLEMS

Example 6.2. Three black and six white beads are strung onto a circular wire. This necklace can be rotated in its plane, and turned over. How many different types of necklaces can be made assuming that beads of the same color are indistinguishable?

Solution. Position the three black and six white beads at the vertices of a regular 9-gon. If the 9-gon is fixed, there are 9 · 8 · 7/3! = 84 different ways of doing this. Two such arrangements are equivalent if there is an element of the symmetry group of the regular 9-gon, D9, which takes one arrangement into the other. The group D9 permutes the different arrangements, and the number of nonequivalent arrangements is the number, N , of orbits under D9. We can now use the Burnside theorem to find N .

Table 6.2 lists all the different types of elements of D9 and the number of fixed points for each type. For example, consider the reflection g D9 about the line joining vertex 2 to the center of the circle, which is illustrated in Figure 6.1. Then the arrangements that are fixed under g occur when the black beads are at vertices 1 2 3, 9 2 4, 8 2 5, or 7 2 6. Hence |Fix g| = 4. There are nine reflections about a line, one through each vertex. Therefore, the total number of fixed points contributed by these types of elements is 9 · 4 = 36. A rotation of order 3 in D9 will fix an arrangement if the black beads are at vertices 1 4 7, 2 5 8, or 3 6 9; hence there are three arrangements that are fixed. If an arrangement is fixed under a rotation of order 9, all the beads must be the same color; hence |Fix g| = 0 if g has order 9. Table 6.2 shows that the sum of all the numbers of

In this example, it is easy to determine all the seven types. They are illustrated in Figure 6.2.

TABLE 6.2. Action of D9 on the Necklaces

Figure 6.2. Seven types of necklaces.

Figure 6.3. Benzene ring.

TABLE 6.3. Action of D6 on the Compounds

Example 6.5. In how many ways is it possible to color a regular dodecahedron so that five of its faces are black and the other seven are white?

Solution. The number of ways† of choosing five faces of a fixed dodecahedron

The different types of elements in the rotation group, A5, of the dodecahedron are shown in Figure 6.5. The numbers of elements of a given type, in Table 6.5, are calculated as follows. An element of order 3 is a rotation about an axis through opposite vertices. Since there are 20 vertices, there are ten such axes. There are two nonidentity rotations of order 3 about each axis; thus the total number of elements of order 3 is 10 · 2 = 20. The elements of orders 2 and 5 can be counted in a similar way.

If g2 A5 is of order 3, we can calculate |Fix g2| as follows. The element g2 does not fix any face and permutes the faces in disjoint 3-cycles. Now five black faces cannot be permuted by disjoint 3-cycles without fixing two faces, so |Fix g2| = 0. Similarly, |Fix g1| = 0 if g1 is a 2-cycle. If g3 is of order 5, then g3 is a rotation about an axis through the centers of two opposite faces, and these two faces are fixed. The other ten faces are permuted in two disjoint 5-cycles; either of these 5-cycles can be black; thus |Fix g3| = 2.

It follows from Table 6.5 and from the Burnside theorem that the number of different colorings is 840/60 = 14.

Any face coloring of the dodecahedron corresponds to a vertex coloring of its dual, the icosahedron.

coefficient.