# Pólya Enumeration Theorem and Chemical Compounds

How many different chemical compounds can be made by attaching H, CH3, or OH radicals to each of the carbon atoms in the benzene ring of Figure 1? (Assume that all of the C–C bonds in the ring are equivalent.)

Figure 1:

I understand the overall method of the Pólya-Burnside Method of Enumeration but I don't understand how to apply it to a chemical compound problem like this. Any help would be great, thank you in advance!

Here is a solution via the Polya Enumeration Theorem. The problem is the same as determining the number of ways of coloring the vertices of a hexagon with three colors.

The symmetry group of the hexagon is the dihedral group $$D_6$$. The cycle index of the group is $$Z=\frac{1}{12} (x_1^6 + 4 x_2^3 +2 x_3^2 +2 x_6 + 3x_1^2 x_2^2)$$ The figure inventory for three colors (or three radicals) is $$x+y+z$$. Substituting the figure inventory into the cycle index yields $$\frac{1}{12}[ (x+y+z)^6 + 4(x^2+y^2+z^2)^3 +2(x^3+y^3+z^3)^2 + 2(x^6+y^6+z^6) + 3(x+y+z)^2(x^2+y^2+z^2)^2 ]$$ If we were to expand this polynomial we could find all the possible ways of coloring the vertices with specific combinations of colors, such as one red vertex, two blue vertices, and three green vertices. But all we really want to know for this problem is the total number of ways with all possible combinations of three colors, which is the sum of the coefficients, and we can find that sum by setting $$x=y=z=1$$, with result $$\frac{1}{12}(3^6 + 4 \cdot 3^3 + 2 \cdot 3^2 + 2 \cdot 3 + 3 \cdot 3^2 \cdot 3^2) = \boxed{92}$$

This is equivalent to the answer by @awkward, so I'm offering this merely as a more "combinatorial" visualization of how to apply Burnside's Lemma.

Following tradition I will talk of coloring a vertex of the hexagon, instead of attaching a radical to a carbon atom.

Of the $$12$$ symmetries of the hexagon:

• The identity element leaves all $$3^6$$ colorings unchanged.

• Rotate-by-$$1$$ and rotate-by-$$5$$ leave a coloring unchanged iff all $$6$$ vertex colors are the same; there are $$3$$ such colorings (i.e. there is free choice of $$1$$ color, which must be used in the entire hexagon).

• Rotate-by-$$2$$ and rotate-by-$$4$$ leave a coloring unchanged iff the vertex colors alternate; there are $$3^2$$ such colorings (i.e. there is free choice of $$2$$ colors, allowing the same color chosen twice).

• Rotate-by-$$3$$ leaves a coloring unchanged iff the colors go $$abcabc$$ around the hexagon; there are $$3^3$$ such colorings.

• There are $$3$$ reflections along a main diagonal, and each will leave a coloring unchanged iff the the off-diagonal vertices are paired in colors (the on-diagonal vertices can be any colors); there are $$3^4$$ such colorings.

• Finally there are $$3$$ reflections along a line bisecting opposite edges, and each will leave a coloring unchanged if the vertex colors form $$3$$ pairs; there are $$3^3$$ such colorings.

So by the magical :) Burnside Lemma, the answer is:

$${1 \over 12} (3^6 + 2 \times 3 + 2 \times 3^2 + 3^3 + 3 \times 3^4 + 3 \times 3^3) = 92$$

Call the radicals $$X$$, $$Y$$ and $$Z$$ for simplicity.

Case 1: all the same (XXXXXX, YYYYYY or ZZZZZZ)

one way for each case:

Total for Case 1 = 3 ways

Case 2: just two radicals used, e.g., (X and Y) or (X and Z) or (Y and Z)

For each pair of radicals used:

5X 1Y: one way

4X 2Y: three ways

3X 3Y: four ways

2X 4Y: three ways

1X 5Y: one way

total for each pair: 12 ways.

Three choices for pairs so...

Total for Case 2 = 36 ways

Case 3: Three radicals: X and Y and Z

subcase a: 2 X, 2 Y, 2 Z

xxx ways

subcase b: 1 X

[I hope you get the principle and can complete this]

In the third case you must use all three radicals, X and Y and Z.

Either each is represented twice (XX and YY and ZZ)

...or not. One of the radicals must appear just once. Suppose it is X.

Then you have:

1Y 4Z

2Y 3Z

3Y 2Z

4Y 1Z

for each of these you can enumerate the distinct cases.