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Chapter 1: Problem 43

In how many ways may we string \(n\) distinct beads on a necklace without a clasp? (Perhaps we make the necklace by stringing the beads on a string, and then carefully gluing the two ends of the string together so that the joint can't be seen. Assume someone can pick up the necklace, move it around in space and put it back down, giving an apparently different way of stringing the beads that is equivalent to the first.) \((\mathrm{h})\)

Short Answer

Expert verified
The number of unique ways to string the beads is \( \frac{(n-1)!}{2} \).

Step by step solution

01

- Understand the problem

The problem is about finding the number of unique ways to arrange distinct beads on a necklace. Since the necklace has no clasp and can be rotated or flipped, two arrangements that look different might be considered the same.
02

- Consider all possible linear arrangements

There are \( n! \) ways to arrange the n distinct beads in a straight line.
03

- Account for rotation

Since rotating the necklace by any number of beads gives the same necklace, we divide by \( n \). Thus, there are \( \frac{n!}{n} \) ways to arrange the beads considering rotations, which simplifies to \( (n-1)! \).
04

- Account for reflection

Further, flipping the necklace over gives the same arrangement, so we must divide by 2 to account for this symmetry. The total number of unique arrangements is thus \( \frac{(n-1)!}{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Necklace Permutations
A necklace permutation refers to the different ways to arrange a set of beads on a circular string where the starting point doesn't matter. In other words, it considers arrangements that are identical when rotated or flipped. For a necklace without a clasp, these permutations are considered the same if one can rotate or reflect the necklace to look identical. This makes the problem more complex than simple linear permutations.
Rotational Symmetry
Rotational symmetry means that you can rotate a shape around a central point and it will look the same from various angles. For a necklace, if you move beads in a circular pattern, the arrangement remains unchanged. This affects permutation calculations because you won't count rotations as distinct. Imagine rotating a 5-bead necklace: every rotation by one bead step results in the same arrangement. This reduces the total unique arrangements. Mathematically, we account for this by dividing by n, the number of beads, turning the calculation from n! to \([\frac{n!}{n}]\), simplifying to \((n-1)!\).
Reflection Symmetry
Reflection symmetry involves flipping an object to see if it remains identical. For a necklace, flipping it over will often produce the same visual arrangement. This removes more unique permutations from consideration. When we reflect an arrangement, we essentially halve the number of distinct permutations left after accounting for rotations. We do this mathematically by dividing the previous result of \((n-1)!\) by 2. This means our total is \([\frac{(n-1)!}{2}]\).
Distinct Arrangements
Distinct arrangements refer to unique patterns you can form without repeating due to rotations or reflections. Calculating these patterns for a necklace gives a more accurate count of unique necklaces since we're considering all symmetrical reductions. The final formula \([\frac{(n-1)!}{2}]\) ensures we're only counting genuinely distinct arrangements. This comprehensive approach accounts for both types of symmetry: rotational and reflectional, ensuring that all counted arrangements are unique.

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Most popular questions from this chapter

If we make a sequence of \(m\) choices for which \- there are \(k_{1}\) possible first choices, and \- for each way of making the first \(i-1\) choices, there are \(k_{i}\) ways to make the \(i\) th choice, then in how many ways may we make our sequence of choices? (You need not prove your answer correct at this time.)

The binary representation of a number \(m\) is a list, or string, \(a_{1} a_{2} \ldots a_{k}\) of zeros and ones such that \(m=a_{1} 2^{k-1}+a_{2} 2^{k-2}+\cdots+a_{k} 2^{0}\) Describe a bijection between the binary representations of the integers between 0 and \(2^{n}-1\) and the subsets of an \(n\) -element set. What does this tell you about the number of subsets of an \(n\) -element set?

(This becomes especially relevant in Chapter \(6,\) though it makes an important point here.) In how many ways may we attach two identical red beads and two identical blue beads to the corners of a square (with one bead per corner) free to move around in (three-dimensional) space? (n)

A roller coaster car has \(n\) rows of seats, each of which has room for two people. If \(n\) men and \(n\) women get into the car with a man and a woman in each row, in how many ways may they choose their seats?

Two sets are said to be disjoint if they have no elements in common. For example, \(\\{1,3,12\\}\) and \(\\{6,4,8,2\\}\) are disjoint, but \(\\{1,3,12\\}\) and \(\\{3,5,7\\}\) are not. Three or more sets are said to be mutually disjoint if no two of them have any elements in common. What can you say about the size of the union of a finite number of finite (mutually) disjoint sets? Does this have anything to do with any of the previous problems?
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