When you do mathematics you spend much of your time being confused.

We can go further with our counting and count the number of necklaces and bracelets with a particular colouring.

When we counted the number of different necklaces using rotational symmetries we arrived at one number and when we added an equal number of reflectional symmetries we counted a lower number.

What happened?

Since we are using small numbers we can do the count by hand. When the numbers grow we will need a more efficient method.

We counted the number of different necklaces and bracelets that we could obtain from \(n\) beads using \(a\) as a variable. We made use of the superscripts (the number of cycles) of \(a\) and ignored the subscripts (the length of the cycles).

When \(a = 1\), we have seen that the count is trivial; we have one necklace or bracelet.

For \(a = 2\), we saw the count go down from \(14\) to \(13\).

We can consider \(a = 2\) as representing red beads and blue beads.

We had \(1/6(a_1^6 + a_2^3 + 2a_3^2 + 2a_6)\) for six (count the coefficients) rotational symmetries giving a count of fourteen different necklaces.

Let us make use of the subscripts in the following way with the help of Wolfram Alpha for the expansions.

Let \(a = 2\) be represented as \(r + b\).

We now use the subscripts as superscripts and expand and add.

\(a_1^6\) becomes \((r + b)^6\).

\(a_2^3\) becomes \((r^2 + b^2)^3\).

\(2a_3^2\) becomes \(2(r^3 + b^3)^2\)

\(2a_6\) becomes \(2(r^6 + b^6)\).

This is :

\(6r^6 + 6r^5b + 18r^4b^2 + 24r^3b^3 + 18r^2b^4 + 6rb^5 + 6b^6\).

Dividing through by six gives :

\(r^6 + r^5b + 3r^4b^2 + 4r^3b^3 +3r^2b^4 + rb^5 + b^6\).

In this way, we are given the information that we have one necklace with six red beads, one with five red and one blue, three with four red and two blue, four with three red and three blue, three with two red and four blue, one with one red and five blue and one with six blue.

Now consider the reflections that we found.

We have :

\(3(r + b)^2(r^2 + b^2)^2 + 3(r^2 + b^2)^3\)

to give :

\(6r^6 + 6r^5b +18r^4b^2 + 12r^3b^3 + 18r^2b^4 + 6rb^5 + 6b^6\)

Dividing through by six gives :

\(r^6 + r^5b + 3r^4b^2 +2r^3b^3 + 3r^2b^4 + rb^5 + b^6\)

We must be very careful at this point because the reflections on their own do not constitute a group because they do not contain the identity element, \(e\).

Nonetheless, we can obtain information from this count. It tells us that of the four necklaces with three red beads and three blue beads having rotational symmetry, only two of them also have reflectional symmetry.

Now for the confusion.

We add the necklaces (rotational symmetry) with the reflections to obtain the number of bracelets (rotations and reflections). This constitutes a group because we have the identity element back. After dividing through by two we have :

\(r^6 + r^5b + 3r^4b^2 +3r^3b^3 + 3 r^2b^4 + rb^5 + b^6\)

Which three of the four necklaces are you going to choose as bracelets?

One and two seem obvious choices since the both have rotational and reflectional symmetry. Of the other two, which one will you choose? Both are equally valid or invalid but only one counts.

A pair of footprints. Neither footprint has reflectional symmetry although both have rotational symmetry. However, there is a vertical axis of symmetry which will reflect one footprint onto the other. This is chirality.

Necklaces three and four are both chiral objects and can be reflected onto each other. Are we treating this pair of necklaces as one bracelet?

When \(a = 3\) we can denote \(a\) as \(r + b + g\).

The rotations (necklaces) are :

\((r + b +g)^6 + (r^2 + b^2 + g^2)^3 +2(r^3 + b^3 + g^3)^2 + 2(r^6 + b^6 + g^6)\).

Expanding, adding and dividing through by six gives :

\(b^6 + b^5g + b^5r \\

+ 3b^4g^2 + 5b^4gr + 3b^4r^2 \\

+ 4b^3g^3 + 10b^3g^2r + 10b^3gr^2 + 4b^3r^3 \\

+ 3b^2g^4 + 10b^2g^3r +16b^2g^2r^2 + 10b^2gr^3 + 3b^2r^4 \\

+ bg^5 + 5bg^4r +10bg^3r^2 +10bg^2r^3 +5bgr^4 + br^5 \\

+ g^6 + g^5r +3g^4r^2 + 4g^3r^3 +3g^2r^4 + gr^5 +r^6\)

for a total of \(130\) necklaces where we can see how many have a particular colouring.

The reflections are :

\(3(r + b + g)^2(r^2 + b^2 +g^2)^2 +3(r^2 + b^2 + g^2)^3\)

Expanding, adding and dividing through by six gives :

\(b^6 + b^5g + b^5r \\

+ 3b^4g^2 + b^4gr + 3b^4r^2 \\

+ 2b^3g^3 + 2b^3g^2r + 2b^3gr^2 + 2b^3r^3 \\

+ 3b^2g^4 +2b^2g^3r + 6b^2g^2r^2 + 2b^2gr^3 + 3b^2r^4 \\

+ bg^5 + bg^4r + 2bg^3r^2 + 2bg^2r^3 +bgr^4 +br^5 \\

+ g^6 + g^5r + 3g^4r^2 + 2g^3r^3 + 3g^2r^4 + gr^5 + r^6\)

for \(54\) necklaces which also have reflectional symmetry.

Adding the two together and dividing by two we find \(92\) bracelets.

\(b^6 + b^5g + b^5r \\

+ 3b^4g^2 + 3b^4gr + 3b^4r^2 \\

+ 3b^3g^3 + 6b^3g^2r + 6b^3gr^2 + 3b^3r^3 \\

+ 3b^2g^4 +6b^2g^3r + 11b^2g^2r^2 + 6b^2gr^3 + 3b^2r^4 \\

+ bg^5 + 3bg^4r + 6bg^3r^2 + 6bg^2r^3 + 3bgr^4 + br^5 \\

+ g^6 + g^5r + 3g^4r^2 +3g^3r^3 + 3g^2r^4 + gr^5 + r^6 \)

Where have \(38\) necklaces gone?

© OldTrout \(2018\)

**No Audio file – it does not translate well**