# Questions tagged [necklace-and-bracelets]

In combinatorics, a *necklace* of length $n$ is an equivalence class of strings of length $n$, under rotation, so $abcde = bcdea$. A *bracelet* is an equivalence class of strings under rotation and reflection (so in addition $abcde = edcba$).

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Consider a circular necklace with $18$ identical beads. We can rotate the necklace and turn it over. Let $G$ denote the symmetry group. How many rotations does $G$ contain? How many reflections does $... 3 votes 2 answers 249 views ### How many necklaces can be formed with$6$identical diamonds and$3$identical pearls Find number of ways to make a necklace (or a garland) consisting of$6$identical diamonds and$3$identical pearls. I got the correct answer$7$by taking different cases but when I applied the ... 2 votes 1 answer 110 views ### How many necklaces made of black and white beads (k total, x black) have at least y consecutive black beads? Consider all necklaces consisting of black and white beads, of length k, containing x black beads. How many such necklaces contain y consecutive black beads somewhere in the necklace? (y is less than ... 1 vote 3 answers 122 views ### Quantifying the evenness-of-distribution of nodes within a necklace Given a necklace with n nodes that are distributed around a circle by a set of given deltas: How would you quantify how evenly the nodes are distributed. (By "evenly" I mean that each node ... 3 votes 1 answer 91 views ### How many necklaces with given certain colored beads I am interested to know if there's a way to calculate the number of (rotation agnostic) necklaces that can be produced from different colored beads, each color with its own quantity. For instance, if ... 1 vote 1 answer 128 views ### How many necklaces are there with a known number of beads of each color? [duplicate] I suspect that the answer to my question might be trivially found in the Wikipedia page for the combinatorial concept of a necklace, but I'm finding that page very hard to understand. Suppose I have$...
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How many different seven-bead necklaces are possible, assuming each bead is one of four different colors and each necklace contains exactly one bead of one color and exactly two beads of the three ...
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### How many different circular necklaces containing ten beads can be made using beads of at most two colors?

How many different circular necklaces containing ten beads can be made using beads of at most two colors? I know I need calculate situation with 2 colors, 8 second color, 3 first color, 7 second ...
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### Find the number of different types of circular necklaces that could be made from the sets of beads

Find the number of different types of circular necklaces that could be made from the sets of beads 7 black and 5 white beads We need to solve this by the Polya-Burnside method of enumeration: Since ...
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### Number of ways to arrange objects in a circle, some of which may be identical

I know that the number of ways to arrange $n$ distinct objects in a circle in $(n-1)!$ from Circular Permutation. But suppose we have $n_1$ identical objects of Type $1$, $n_2$ identical objects of ...
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### Using burnside's lemma to calculate a smaller subset of unique, color-agnostic bracelets

We have a child's toy, which is a ball made of 12 colored wedges (3 Red, 3 Green, 3 Blue, 3 Yellow). Our child asked the sensible question 'how many different patterns are possible?'. In researching ...
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### Finding "beautiful" necklaces with regular gaps

I am looking for "beautiful" arrangements of $k$-ary necklaces of length $ak$ where each of the $k$ types of bead appears $a$ times ($a \geq 1$ a natural number). A necklace is considered ...
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### Rotation of necklaces

The number of fixed necklaces of length $n$ with $a$ types of beads is $$N(n,a)=\frac1n\sum_{d|n}\phi(d)a^{n/d}\;.$$ It is clear intuitively that the number of rotational coincidences gets ...
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### (Music) List of all possible "types of set" of 12 musical notes

I have looked into trying to figure what are all the possible "types" of note set combinations there are and how I would go by listing them if possible. It turns out this is harder than I thought. The ...