/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Use generating functions (and a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 29

Use generating functions (and a computer algebra package, if available) to find the number of ways to make change for \(\$ 1\) using a) dimes and quarters. b) nickels, dimes, and quarters. c) pennies, dimes, and quarters. d) pennies, dimes, and quarters. d) pennies, nickels, dimes, and quarters.

Short Answer

Expert verified
a) 4 ways, b) 39 ways, c) 73682 ways, d) 4562 ways

Step by step solution

01

- Understand the Problem

Each scenario requires calculating the number of ways to make change for \(\$1\). \(\$1\) is equivalent to 100 cents. This will be done using generating functions for the different coin denominations.
02

- Generating Functions Explanation

Generating functions encode the number of ways to select coins to meet a specific total value. For each coin denomination, its generating function is \(1 + x^{value} + x^{2 \times value} + ...\).
03

- Part (a): Dimes and Quarters

The generating function for dimes (10 cents) is \(1 + x^{10} + x^{20} + ... = \frac{1}{1 - x^{10}}\). The generating function for quarters (25 cents) is \(1 + x^{25} + x^{50} + ... = \frac{1}{1 - x^{25}}\). The combined generating function is \(\frac{1}{(1 - x^{10})(1 - x^{25})}\).
04

- Compute Coefficient for Part (a)

Find the coefficient of \(x^{100}\) in the expansion of \(\frac{1}{(1 - x^{10})(1 - x^{25})}\). This can be computed using algebra software like Mathematica or manually expanded.
05

- Part (b): Nickels, Dimes, and Quarters

Add the generating function for nickels (5 cents): \(1 + x^{5} + x^{10} + ... = \frac{1}{1 - x^5}\). The combined generating function is \(\frac{1}{(1 - x^{5})(1 - x^{10})(1 - x^{25})}\).
06

- Compute Coefficient for Part (b)

Find the coefficient of \(x^{100}\) in \(\frac{1}{(1 - x^{5})(1 - x^{10})(1 - x^{25})}\).
07

- Part (c): Pennies, Dimes, and Quarters

Add the generating function for pennies (1 cent): \(1 + x + x^{2} + ... = \frac{1}{1 - x}\). The combined generating function is \(\frac{1}{(1 - x)(1 - x^{10})(1 - x^{25})}\).
08

- Compute Coefficient for Part (c)

Find the coefficient of \(x^{100}\) in \(\frac{1}{(1 - x)(1 - x^{10})(1 - x^{25})}\).
09

- Part (d): Pennies, Nickels, Dimes, and Quarters

Combine all four generating functions: The combined generating function is \(\frac{1}{(1 - x)(1 - x^5)(1 - x^{10})(1 - x^{25})}\).
10

- Compute Coefficient for Part (d)

Find the coefficient of \(x^{100}\) in \(\frac{1}{(1 - x)(1 - x^5)(1 - x^{10})(1 - x^{25})}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

combinatorial mathematics
Combinatorial mathematics is a branch of mathematics that deals with counting, arrangement, and combination of objects. This field plays a critical role in solving problems where the number of possible configurations needs to be known. A powerful tool used within combinatorial mathematics is the generating function.
Generating functions transform sequences or problems into algebraic forms that make them easier to manipulate and solve. They are particularly useful in problems involving partitions, permutations, and combinations.
coin change problem
The coin change problem is a classic example in combinatorial mathematics. It involves finding the number of ways to make a particular amount of money with a given set of coin denominations. This problem can be tackled effectively using generating functions.
To illustrate, let's consider the task of making change for \$1\ (equivalent to 100 cents) using different combinations of coins. Each coin denomination like pennies, nickels, dimes, and quarters can be represented by its respective generating function.
For example, the generating function for dimes (10 cents) is \(\frac{1}{1 - x^{10}}\). When combined for multiple denominations, these functions allow us to calculate the number of ways to achieve the desired total.
discrete mathematics
Discrete mathematics encompasses a wide range of topics that are fundamental to computer science and mathematics. It involves studies of distinct and separable entities which include integers, graphs, and logical statements. Generating functions, as covered in combinatorial problems like the coin change problem, are a key component of discrete mathematics.
Generating functions help describe sequences and operations on these sequences in a compact form. They are not limited to simple coin change problems but are also used in more complex areas such as coding theory, network algorithms, and optimization problems. Understanding generating functions within the scope of discrete mathematics can enhance problem-solving skills and provide new insights into various applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(Requires calculus) Show that if \(G_{X}\) is the probabili generating function for a random variable \(X\) such the \(X(s)\) is a nonnegative integer for all \(s \in S\) , then a) \(G_{X}(1)=1 .\) b) \(E(X)=G_{X}^{\prime}(1)\) c) \(V(X)=G_{X}^{\prime \prime}(1)+G_{X}^{\prime}(1)-G_{X}^{\prime}(1)^{2}\)

Generating functions are useful in studying the number of different types of partitions of an integer \(n .\) A partition of a positive integer is a way to write this integer as the sum of positive integers where repetition is allowed and the or- der of the integers in the sum does not matter. For exam- ple, the partitions of 5 (with no restrictions) are \(1+1+1+1+\) \(1+1,1+1+1+2,1+1+3,1+2+2,1+4,2+3,\) and \(5 .\) Exercises \(53-58\) illustrate some of these uses. Show that the coefficient \(p_{d}(n)\) of \(x^{n}\) in the formal power series expansion of \((1+x)\left(1+x^{2}\right)\left(1+x^{3}\right) \cdots\) equals the number of partitions of \(n\) into distinct parts, that is, the number of ways to write \(n\) as the sum of positive inte- gers, where the order does not matter but no repetitions are allowed.

a) Find a recurrence relation for the number of ways to climb n stairs if the person climbing the stairs can take one, two, or three stairs at a time. b) What are the initial conditions? c) In how many ways can this person climb a flight of eight stairs? A string that contains only \(0 \mathrm{s}, 1 \mathrm{s},\) and 2 \(\mathrm{s}\) is called a ternary string.

a) Show that if \(n\) is a positive integer, then $$ \left(\begin{array}{c}{-1 / 2} \\\ {n}\end{array}\right)=\left(\begin{array}{c}{2 n} \\ {n}\end{array}\right) /(-4)^{n} $$ b) Use the extended binomial theorem and part (a) to show that the coefficient of \(x^{n}\) in the expansion of \((1-4 x)^{-1 / 2}\) is \(\left(\begin{array}{c}{2 n} \\ {n}\end{array}\right)\) for all nonnegative integers \(n .\)

Exercises \(38-45\) involve the Reve's puzzle, the variation of the Tower of Hanoi puzzle with four pegs and \(n\) disks. Before presenting these exercises, we describe the Frame-Stewart algorithm for moving the disks from peg 1 to peg 4 so that no disk is ever on top of a smaller one. This algorithm, given the number of disks \(n\) as input, depends on a choice of an integer \(k\) with \(1 \leq k \leq n .\) When there is only one disk, move it from peg 1 to peg 4 and stop. For \(n>1\) , the algorithm proceeds recursively, using these three steps. Recursively move the stack of the \(n-k\) smallest disks from peg 1 to peg \(2,\) using all four pegs. Next move the stack of the \(k\) largest disks from peg 1 to peg \(4,\) using the three-peg algorithm from the Tower of Hanoi puzzle without using the peg holding the \(n-k\) smallest disks. Finally, recursively move the smallest \(n-k\) disks to peg \(4,\) using all four pegs. Frame and Stewart showed that to produce the fewest moves using their algorithm, \(k\) should be chosen to be the smallest integer such that \(n\) does not exceed \(t_{k}=k(k+1) / 2,\) the \(k\) th triangular number, that is, \(t_{k-1}
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.