Math 471
Group Work #13
SOLUTIONS
Fall 2014
Problem 1. A merry-go-round can be built with three different styles of horses.
How many five-horse merry-go-rounds are there?
SOLUTION: We can model the merry-go-round as a pentagon, each of whose vertices
can be colored one of 3 colors, and which can be rotated but not flipped. So the group G
of symmetries of the merry-go-round are the 5 rotations of a pentagon. The identity
permutation fixes all 35 colorings of the corners of the pentagon, while each of the other 4
rotations of the pentagon will only fix colors that consist of all vertices colored the same.
There are only 3 such colorings. Hence, by Burnside’s Theorem, the number of fivehorse merry-go-rounds is
(1/5)[35 + 3 + 3 + 3 + 3] = 255/5 = 51.
Problem 2. A baton is painted with equal-sized cylindrical bands. Each band can
be painted red, white, or blue. If the band is unoriented as when spun in the air,
how many different 3-colorings of the baton are possible if the baton has 4 bands? 5
bands?
SOLUTION: A baton can be modeled as a graph which is simply a straight line with n
vertices equally-spaced along the line to represent the n bands. Aside from the identity
symmetry, there is only one other symmetry, which is to rotate (or flip) the baton along
the perpendicular bisector of the baton.
For 4 bands, there are 34 colorings, all of which are fixed by the identity symmetry, and
32 of them are fixed by the flip. So by Burnside’s Theorem, there are (1/2)[81+9] = 45
different 3-colorings of a baton with 4 bands.
For 5 bands, there are 35 colorings, all of which are fixed by the identity symmetry, and
33 of them are fixed by the flip. So by Burnside’s Theorem, there are (1/2)[243+27] =
135 different 3-colorings of a baton with 5 bands.
Problem 3. Find the number of different p-colorings of the vertices of each of the
following floating figures:
(a)
SOLUTION: There are only two symmetries of the figure (the identity symmetry and
the flip along the central horizontal axis). The first symmetry fixes all p5 colorings, while
the latter symmetry fixes p3 colorings. By Burnside’s Theorem, there are
(1/2)[p5 + p3] colorings.
(b)
SOLUTION: There are four symmetries of this figure (identity, 180-rotation, flip along
horizontal axis, and flip along vertical axis). The number of colorings fixed by these
symmetries are, respectively, p8,p4, p4 ,p4. By Burnside’s Theorem, there are
(1/4)[p8 +3 p4] colorings.
(c)
SOLUTION: This figure enjoys the same 8 symmetries as the square. I’m going to
leave it to you to check that the answer here is (1/8)[p12 + 2p3 + 3p6 + 2p7].