How many arrangements of beads are possible in a bracelet if there are 6 different designs of beads?

Suppose you’re making a necklace of 10 beads, and you have 3 kinds of beads: red, green and blue. How many different necklaces can you make?

How many of these necklaces contain at least 3 red beads? How many are composed of 2 red beads, 3 green beads and 5 blue beads?

At first glance, these may not seem that tricky to answer. After all, consider the similar question: how many ways can you arrange 10 different beads in a necklace? This is easy: count all permutations of 10 beads, 10!, then divide by 20 because we counted each permutation 10 times due to rotation, and counted each of these twice because you can flip the necklace over. Thus the answer is 10!/20 = 181440.

Perhaps we can solve the first problem with a similar technique. Let’s try a smaller case for now and study 6-bead necklaces. The first step is easy: the number of ways to colour 6 beads, where each bead can be red, green or blue, is 36 = 729. Next we put the beads on a necklace, and account for duplicate patterns. For example, RRRRRG and RRRRGR are the same pattern, because you can rotate the necklace to go from one to the other. On the other hand, some patterns such as RRRRRR only get counted once.

One might hope there are only a few different cases so we can tweak the total a little to get the right answer. But the necklace consisting of 5 red beads and 1 green bead appears 6 times, the necklace consisting of all red beads appears once, the necklace consisting of 4 red beads and 2 green beads at opposite ends such as RRGRRG appears 3 times, and so on. We haven’t even considered blue beads yet!

Furthermore, dramatic changes occur when only one more bead is added. For instance, there are 7 colourings equivalent to RRGRRRG, and in fact, each colouring for the 7 bead case will appear either 1, 2, 7, or 14 times.

There’s no easy way out. To answer the question, we must study the symmetry group of the n-bead necklace: the dihedral group of order 2n.

Why doesn’t the symmetry group matter for the easy variant of the question, where each bead is unique? Actually, it does, but the only information we need is the size of the symmetry group. When each bead is unique, different rotation and reflection operations are guaranteed to produce different colourings, so we know we counted each one exactly 20 times.

When beads can have the same colour as other beads, rotations or reflections can leave a colouring unchanged. For example, rotating RRGRRG by 3 beads has no effect. Our counting algorithm must therefore intimately involve the symmetry group.

References

See K. H. Wehrhahn, "Combinatorics: An Introduction".

Answer:

= 720 distinct ways (permutations) to arrange the 6 different beads. With ends attached to form what is essentially a circle, the multiplier 6 in 6! collapses to 1 and the number of distinct arrangements of the beads becomes just (6/6)(5!) = 5!

Step-by-step explanation:

Hope It Helps

Research Makes Perfect

Paul Raff gave a formula for both bracelets and necklaces so in my answer, I will provide a general method that you can use for this kind of problem. It works also if you want to colour a cube for example.

As Paul Raff pointed out, you did get mix up between bracelet and necklace so in my answer I will include the answer for both of them.

Where did you get it wrong?

It seems to me that you are counting number of ways to colour a bracelet rather than a necklace so I checked your calculation with respect to colouring a bracelet. I would say that the main problem in your counting is that even for each case of the base, you cannot always guarantee to get the final result for that case by multiplying as you did. For example, in the case of $A^2B^2C^2$, we consider two following iterations:

The one on the left gives $\frac{6\cdot5\cdot 4}{3!}=\frac{120}{6}$ ways to choose three colours $A,B,C$, which is what you gave in your table. However, the one on the right gives $\frac{6 \cdot 5 \cdot 4}{2}=\frac{120}{2}$ ways to choose three colours $A,B,C$.

Necklace colouring

You can use Burnside lemma where you can count the number of ways to colour the object by looking at its group of symmetry $G$. For the necklace, the group $G$ can be:

• Two colourings of the necklace are considered the same if from one colouring, we can rotate the necklace to get to the other colouring. There are $5$ possible rotations at angles $60^{\circ}\cdot i \; (i=1,2,3,4,5)$ (not including the do-nothing rotation). Hence, these five rotations are from the mentioned set $G$.
• The "do nothing" action, i.e. we do nothing to the necklace. This is also in $G$.

Let $X$ be the set of all possible colouring for the necklace at a fixed orientation. This follows $|X|=6^6$ as there are $6$ possible colours for each bead.

Now, in Burnside lemma, we essentially want to count number of colourings from $X$ that remains unchanged under actions from $G$. In particular:

• How many colourings in $X$ for the necklace so that it is still the same colouring after we apply $60^{\circ}$ rotation to the necklace? This only happens when all the beads have the same colour. Hence, there are $6$ possible colourings in this case.

• The same question can be asked for $120^{\circ}$ rotation: This happens when three of the beads (each is one bead away from the other) have the same colour and the remaining three beads have the same colour. There are $6$ possible colours for the first three beads and there are $6$ other possible colours for remaining $6$ beads. This gives us $6^2$ possible colourings.

• Similarly, with $180^{\circ}$ rotation, a colouring is fixed under this rotation when any pair of opposite beads have the same colour. There are $6$ ways to colour each pair of opposite beads so there are $6^3$ possible colourings.

• With $240^{\circ}$ rotation, it's the same as $120^{\circ}$ so we have $6^2$ possible colourings. With $300^{\circ}$, it's the same as $60^{\circ}$ so we have $6$ possible colourings.

• With the "do nothing" action then every colouring remains unchanged after this action so there are $6^6$ possible colourings.

The Burnside lemma says that you can add all these numbers up and divide by number of elements of $G$ (which is $6$) to obtain all possible colourings. Hence, the answer for colouring a necklace is $$\frac{6^6 \cdot 1 +6 \cdot 2+ 6^2 \cdot 2+6^3 \cdot 1}{6}=7826.$$

Bracelet colouring

The difference between bracelets and necklaces is in the group of symmetry $G$. In particular, for bracelets, $G$ has some extra elements: Two colourings of the bracelet are considered same if from one colouring, we can reflect the bracelet through a line to obtain the other colouring. There are two types of lines:

• A line connecting two opposite beads (see diagram on the right). There are $3$ pairs of opposite beads so there are three reflections through these types of lines. These actions are in $G$.
• A line dividing the $6$ beads in equal parts (diagram on the left). There are $3$ such lines corresponding to three extra reflections in $G$.

This time $G$ has $12$ elements.

Next, we do the same thing with necklace, i.e. we count number of colourings that remains fixed under these reflections:

• For reflection across axis on the left, any two beads that appear symmetrically through that axis must have same colour. There are $3$ pairs of such beads so there are $6^3$ possible colourings. Since there are three axes of this type so we have $3 \cdot 6^3$.
• For reflection across axis on the right, note that we can freely choose any colour for the beads that pass through the axis while keeping the same colouring of the bracelet when reflecting. Hence, this gives $6^2$ for such two beads, and $6^2$ for $2$ pairs of symmetrical beads. We obtain $6^4$ possible colourings for such axis. As there are three axes of this type so it is $3 \cdot 6^4$.

Now, applying Burnside's lemma, we sum up all these numbers counted for each element in $G$ then we divide by number of elements in $G$ (which is $12$). The final answer is $$\frac{1}{12}\left( 6^6 \cdot 1+6 \cdot 2+ 6^2 \cdot 2+6^3 \cdot 1+6^3 \cdot 3+ 6^4 \cdot 3 \right)=4291.$$

How many arrangements of beads are possible in a bracelet if there are 6 different designs of beads illustrate your answer using permutation with complete solution?

How many ways can 5 beads be arranged in a circular bracelet?

How many different kinds of bracelets can be created from 10 different beads?

How many ways can 8 beads be arranged in a bracelet?