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Chapter 2: Problem 44

In the design of an electromechanical product, 12 components are to be stacked into a cylindrical casing in a manner that minimizes the impact of shocks. One end of the casing is designated as the bottom and the other end is the top. (a) If all components are different, how many different designs are possible? (b) If seven components are identical to one another, but the others are different, how many different designs are possible? (c) If three components are of one type and identical to one another, and four components are of another type and identical to one another, but the others are different, how many different designs are possible?

Short Answer

Expert verified
(a) 479,001,600 designs; (b) 95,040 designs; (c) 3,326,400 designs.

Step by step solution

01

Understanding the Problem

We need to determine the number of different ways to arrange components within a cylindrical casing. The problem consists of three parts: (a) Arranging 12 unique components. (b) Arranging 12 components where 7 are identical. (c) Arranging components with different types in certain identical groups.
02

Calculate the Number of Arrangements for (a)

For part (a), all 12 components are unique. The number of ways to arrange them can be calculated using the factorial function: \[ 12! \]This represents the total number of permutations of 12 unique items.
03

Compute 12!

Calculate \(12!\): \[ 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 479001600 \]Thus, there are 479,001,600 possible designs when all components are different.
04

Calculate the Number of Arrangements for (b)

For part (b), 7 components are identical, and the rest are different. The number of arrangements is given by: \[ \frac{12!}{7!} \]This accounts for the identical components which do not contribute to unique permutations.
05

Compute \(\frac{12!}{7!}\)

First, calculate \(7!\): \[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \]Then, compute: \[ \frac{12!}{7!} = \frac{479001600}{5040} = 95040 \]Thus, there are 95,040 designs possible when 7 components are identical.
06

Calculate the Number of Arrangements for (c)

In part (c), 3 components of one type are identical, 4 components of another type are identical, and the remaining are different. The arrangement count is:\[ \frac{12!}{3! \times 4!} \]
07

Compute \(\frac{12!}{3! \times 4!}\)

First, calculate \(3!\) and \(4!\): \[ 3! = 6 \quad \text{and} \quad 4! = 24 \]Next, compute:\[ \frac{12!}{3! \times 4!} = \frac{479001600}{6 \times 24} = \frac{479001600}{144} = 3326400 \]Thus, there are 3,326,400 designs possible with the given identical groups of components.

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Key Concepts

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

Permutations
In combinatorial analysis, permutations are ways to arrange a distinct set of objects. Imagine having a deck of 12 unique cards.
Now, think about all the different possible orders you could lay them out in a line.
Each distinct arrangement represents a unique permutation.This concept helps us answer questions like, "In how many ways can we arrange these 12 components?"
  • If every component is different, the number of permutations is simply the factorial of the number of items, denoted as \( n! \).
    For example, for 12 components, it's \( 12! \).
  • If some components are identical, permutations adjust accordingly, limiting the total different arrangements since swapping identical items doesn't create a new order.
Understanding permutations is crucial for this exercise, which is about evaluating possible designs with given components.
Factorials
The factorial is a crucial concept in permutations. It's a function applied to a number, represented by an exclamation point (!).
For example, \( n! \) is the product of all positive integers less than or equal to \( n \).
It's a straightforward yet powerful tool that allows us to calculate permutations quickly.For instance, \( 3! \) means \( 3 \times 2 \times 1 = 6 \).
  • Factorials grow rapidly.
    Even a small number like 12 results in \( 12! = 479001600 \).
  • They are primarily used to find the number of ways we can order unique items, but they also help us handle constraints, like identical objects.
Using factorials effectively simplifies the complex permutations calculations seen in this example.
Identical Objects
Identical objects in permutation problems add an interesting twist. Imagine you have 12 slots to fill with some objects being indistinguishable.
The concept of identical objects helps us adjust our counting by recognizing that switching two identical objects does not create a new arrangement.To find the permutation with identical objects:
Use the formula \( \frac{n!}{k_1! \times k_2! \times \ldots} \), where \( k_1, k_2, \ldots \) represent the counts of each set of identical items.
  • For 7 identical objects out of 12, the arrangement count is \( \frac{12!}{7!} \).
  • If you have groups of different identical objects, for example, 3 and 4 identical items, you'd use \( \frac{12!}{3! \times 4!} \).
Incorporating identical objects into permutations helps accurately calculate distinct arrangements, making efficient use of combinatorial principles.

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Most popular questions from this chapter

A lot contains 15 castings from a local supplier and 25 castings from a supplier in the next state. Two castings are selected randomly, without replacement, from the lot of \(40 .\) Let \(A\) be the event that the first casting selected is from the local supplier, and let \(B\) denote the event that the second casting is selected from the local supplier. Determine: (a) \(P(A)\) (b) \(P(B \mid A)\) (c) \(P(A \cap B)\) (d) \(P(A \cup B)\) Suppose that 3 castings are selected at random, without replacement, from the lot of \(40 .\) In addition to the definitions of events \(A\) and \(B,\) let \(C\) denote the event that the third casting selected is from the local supplier. Determine: (e) \(P(A \cap B \cap C)\) (f) \(P\left(A \cap B \cap C^{\prime}\right)\)

A Web ad can be designed from four different colors, three font types, five font sizes, three images, and five text phrases. A specific design is randomly generated by the Web server when you visit the site. Determine the probability that the ad color is red and the font size is not the smallest one.

A steel plate contains 20 bolts. Assume that 5 bolts are not torqued to the proper limit. 4 bolts are selected at random, without replacement, to be checked for torque. (a) What is the probability that all 4 of the selected bolts are torqued to the proper limit? (b) What is the probability that at least 1 of the selected bolts is not torqued to the proper limit?

A wireless garage door opener has a code determined by the up or down setting of 12 switches. How many outcomes are in the sample space of possible codes?

An article in the British Medical Journal ["Comparison of treatment of renal calculi by operative surgery, percutaneous nephrolithotomy, and extracorporeal shock wave lithotripsy" (1986, Vol. 82, pp. \(879-892\) ) ] provided the following discussion of success rates in kidney stone removals. Open surgery had a success rate of \(78 \%(273 / 350)\) and a newer method, percutaneous nephrolithotomy (PN), had a success rate of \(83 \%(289 / 350)\). This newer method looked better, but the results changed when stone diameter was considered. For stones with diameters less than 2 centimeters, \(93 \%(81 / 87)\) of cases of open surgery were successful compared with only \(83 \%(234 / 270)\) of cases of PN. For stones greater than or equal to 2 centimeters, the success rates were \(73 \%(192 / 263)\) and \(69 \%(55 / 80)\) for open surgery and PN, respectively. Open surgery is better for both stone sizes, but less successful in total. In \(1951,\) E. H. Simpson pointed out this apparent contradiction (known as Simpson's paradox), and the hazard still persists today. Explain how open surgery can be better for both stone sizes but worse in total.
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