Instruction: What are some practical applications of Burnside's lemma in combinatorics based on this passage?
Context: Burnside's lemma, sometimes also called Burnside's counting theorem, the Cauchy–Frobenius lemma, the orbit-counting theorem, or the lemma that is not Burnside's, is a result in group theory that is often useful in taking account of symmetry when counting mathematical objects. Its various eponyms are based on William Burnside, George Pólya, Augustin Louis Cauchy, and Ferdinand Georg Frobenius. The result is not due to Burnside himself, who merely quotes it in his book 'On the Theory of Groups of Finite Order', attributing it instead to Frobenius (1887).[1] Burnside's Lemma counts "orbits", which is the same thing as counting distinct objects taking account of a symmetry. Other ways of saying it are counting distinct objects up to an equivalence relation R, or counting objects that are in canonical form.

In the following, let G be a finite group that acts on a set X. For each g in G, let Xg denote the set of elements in X that are fixed by g (also said to be left invariant by g), that is, Xg = { x ∈ X | g.x = x }. Burnside's lemma asserts the following formula for the number of orbits, denoted |X/G|:[2]

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{\displaystyle |X/G|={\frac {1}{|G|}}\sum _{g\in G}|X^{g}|.}
Thus the number of orbits (a natural number or +∞) is equal to the average number of points fixed by an element of G (which is also a natural number or infinity). If G is infinite, the division by |G| may not be well-defined; in this case the following statement in cardinal arithmetic holds:

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{\displaystyle |G||X/G|=\sum _{g\in G}|X^{g}|.}
Examples of applications to enumeration
Necklaces
There are 8 possible bit vectors of length 3, but only four distinct 2-colored necklaces of length 3: 000, 001, 011, and 111, because 100 and 010 are equivalent to 001 by rotation; similarly 110 and 101 are equivalent to 011. The formula is based on the number of rotations, which in this case is 3 (including the null rotation), and the number of bit vectors left unchanged by each rotation. All 8 bit vectors are unchanged by the null rotation, and two (000 and 111) are unchanged by each of the other two rotations, giving: 
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{\displaystyle 4={\frac {1}{3}}(8+2+2)}.

For length 4, there are 16 possible bit vectors; 4 rotations; the null rotation leaves all 16 bit vectors unchanged; the 1-rotation and 3-rotation each leave two bit vectors unchanged (0000 and 1111); the 2-rotation leaves 4 bit vectors unchanged (0000, 0101, 1010, and 1111); giving: 
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{\displaystyle 6={\frac {1}{4}}(16+2+4+2)}. These are: 0000, 0001, 0011, 0101, 0111, and 1111.

Colorings of a cube
The number of rotationally distinct colourings of the faces of a cube using three colours can be determined from this formula as follows.

Let X be the set of 36 possible face colour combinations that can be applied to a cube in one particular orientation, and let the rotation group G of the cube act on X in the natural manner. Then two elements of X belong to the same orbit precisely when one is simply a rotation of the other. The number of rotationally distinct colourings is thus the same as the number of orbits and can be found by counting the sizes of the fixed sets for the 24 elements of G.


Cube with coloured faces
one identity element which leaves all 36 elements of X unchanged
six 90-degree face rotations, each of which leaves 33 of the elements of X unchanged
three 180-degree face rotations, each of which leaves 34 of the elements of X unchanged
eight 120-degree vertex rotations, each of which leaves 32 of the elements of X unchanged
six 180-degree edge rotations, each of which leaves 33 of the elements of X unchanged
A detailed examination of these automorphisms may be found here.

The average fix size is thus

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{\frac  {1}{24}}\left(3^{6}+6\cdot 3^{3}+3\cdot 3^{4}+8\cdot 3^{2}+6\cdot 3^{3}\right)=57.
Hence there are 57 rotationally distinct colourings of the faces of a cube in three colours. In general, the number of rotationally distinct colorings of the faces of a cube in n colors is given by

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{\frac  {1}{24}}\left(n^{6}+3n^{4}+12n^{3}+8n^{2}\right).
8 Queens Puzzle
In the eight queens puzzle there are 92 solutions, of which 12 fundamental solutions are distinct up to rotation and reflection of the board. There are 8 combinations of rotations and reflections, including the null action. The null action leaves all 92 solutions unchanged. Four of the 92 solutions are symmetrical, unchanged by 180° rotation. That gives: 
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{\displaystyle 12={\frac {1}{8}}(92+4)}.[3]
Response: * Necklace counting problem: counting rotationally distinct necklaces of various lengths where each bead can be one of the two colors. The article counts necklaces of lengths 3 and 4, and the answer is 4 and 6, respectively.
* Colorings of a cube problem: count rotationally distinct colorings of faces of a cube. The article considers a case when 3 colors are used and the answer is 57.
* 8 queens puzzle: a classical version of the problem is counting different ways to place 8 chess queens onto an 8x8 board so that no queen attacks another (the answer is 92). With Burnside's lemma application the article counts the number of different placements of queens up to rotation and reflection, arriving at 12 as the answer.