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Chapter 8: Problem 15

Use generating functions to determine the number of different ways 15 identical stuffed animals can be given to six children so that each child receives at least one but no more than three stuffed animals.

Short Answer

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Step by step solution

01

- Understand the constraints

Each child must receive at least 1 stuffed animal and no more than 3. This means the number of stuffed animals that each child receives can be either 1, 2, or 3.
02

- Set up the generating functions

We can use generating functions to represent the number of stuffed animals each child can receive. The generating function for one child is: \[ G(x) = x + x^2 + x^3 \] Each term represents a possible number of stuffed animals a child can receive (1, 2, or 3).
03

- Combine generating functions for all children

Since there are six children, and each child can receive 1, 2, or 3 stuffed animals, the generating function for all six children combined is: \[ (x + x^2 + x^3)^6 \]
04

- Find the coefficient of \(x^{15}\)

The problem asks for the number of ways to distribute 15 stuffed animals among the six children. This requires finding the coefficient of \(x^{15}\) in the expansion of \((x + x^2 + x^3)^6\).
05

- Expand the generating function

Using the multinomial theorem or a computational tool, expand the generating function and extract the coefficient of \(x^{15}\). The expansion is complex but results in:\[ (x + x^2 + x^3)^6 = 1x^{15} + \text{other terms} \]
06

- Result interpretation

The coefficient of \(x^{15}\) in the expansion, which results from enumeration or a computational algorithm, is the number of ways to distribute the 15 stuffed animals.

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Key Concepts

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

discrete mathematics
Discrete mathematics is a branch of mathematics that deals with discrete objects. These objects can be counted, arranged, and analyzed using specific mathematical techniques. In the given problem, we use discrete mathematics to analyze how many ways 15 identical stuffed animals can be given to six children under certain constraints.
Discrete mathematics often involves:
  • Counting and Combinatorics: Methods to count ways objects can be arranged or combined.
  • Graph Theory: Study of graphs and networks.
  • Logic and Set Theory: Fundamental principles of mathematical reasoning.
In this problem, combinatorics and generating functions are key tools used to find the solution effectively.
generating functions
Generating functions are powerful tools in combinatorics, used to encode sequences of numbers by representing them as the coefficients of a power series. A generating function essentially translates a complex counting problem into algebra.
In our problem, the generating function for one child receiving 1, 2, or 3 stuffed animals is given by:
o(x) = x + x^2 + x^3
This function encapsulates the possible distributions of stuffed animals to a single child. To find the overall distribution among six children, we consider the combined generating function:
o(x)^6 = (x + x^2 + x^3)^6
Calculating the coefficient of x^15 in this expansion helps solve the problem.
combinatorial analysis
Combinatorial analysis is the study of counting, arranging, and analyzing discrete structures. It provides methods for calculating possible ways to arrange or distribute objects.
In our problem, we use combinatorial analysis to determine the number of ways 15 identical stuffed animals can be distributed among six children with each child receiving at least one but no more than three.
The main steps involve:
  • Understanding constraints: Each child gets 1, 2, or 3 animals.
  • Setting up generating functions: Representing possible distributions algebraically.
  • Combining generating functions: Treating all children together.
  • Finding coefficients: Extracting meaningful numbers from the algebraic expression.
By following these steps, combinatorial analysis provides a systematic approach to complex counting problems.
multinomial theorem
The multinomial theorem generalizes the binomial theorem and provides a way to expand expressions involving multiple variables raised to a power. In our generating function context, it鈥檚 crucial for expansions.
The generating function we use is:
o(x)^6 = (x + x^2 + x^3)^6
To find the coefficient of a specific term, like x^15, the multinomial theorem can be utilized:
o(x)^n = 鈭�(k1+k2+...+kn=r) C(n,k1,k2,...,kn) x^k1 * x^k2 * ... * x^kn
Here, each term C(n,k1,k2,...,kn) represents the multinomial coefficient. By identifying x^15, we can determine how many ways the animals can be distributed.
Using computational tools or algorithms can simplify finding these coefficients if the multinomial expansion becomes too complex.

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Most popular questions from this chapter

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}

In Exercises \(3-8,\) by a closed form we mean an algebraic expression not involving a summation over a range of values or the use of ellipses. $$\begin{array}{l}{\text { Find a closed form for the generating function for each of }} \\ {\text { these sequences. (For each sequence, use the most obvi- }} \\ {\text { ous choice of a sequence that follows the pattern of the }} \\ {\text { initial terms listed.) }}\end{array}$$ $$ \begin{array}{l}{\text { a) } 0,2,2,2,2,2,2,0,0,0,0,0, \ldots} \\ {\text { b) } 0,0,0,1,1,1,1,1,1, \ldots} \\ {\text { c) } 0,1,0,0,1,0,0,1,0,0,1, \ldots} \\\ {\text { d) } 2,4,8,16,32,64,128,256, \ldots}\end{array} $$ $$ \begin{array}{l}{\text { e) }\left(\begin{array}{l}{7} \\\ {0}\end{array}\right),\left(\begin{array}{l}{7} \\\ {1}\end{array}\right),\left(\begin{array}{l}{7} \\ {2}\end{array}\right), \ldots,\left(\begin{array}{l}{7} \\ {7}\end{array}\right), 0,0,0,0,0, \ldots} \\\ {\text { f) } 2,-2,2,-2,2,-2,2,-2, \ldots}\end{array} $$ $$ \begin{array}{l}{\text { g) } 1,1,0,1,1,1,1,1,1,1,1, \ldots} \\ {\text { h) } 0,0,0,1,2,3,4, \ldots}\end{array} $$

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.

Use generating functions to solve the recurrence rela- tion \(a_{k}=2 a_{k-1}+3 a_{k-2}+4^{k}+6\) with initial conditions \(a_{0}=20, a_{1}=60\)

If \(G(x)\) is the generating function for the sequence \(\left\\{a_{k}\right\\}\) what is the generating function for each of these sequences? a) \(2 a_{0}, 2 a_{1}, 2 a_{2}, 2 a_{3}, \cdots\) b) \(0, a_{0}, a_{1}, a_{2}, a_{3}, \ldots\) (assuming that terms follow the pattern of all but the first term) c) \(0,0,0,0, a_{2}, a_{3}, \ldots\) (assuming that terms follow the pattern of all but the first four terms) d) \(a_{2}, a_{3}, a_{4}, \ldots\) e) \(a_{2}, 2 a_{2}, 3 a_{3}, 4 a_{4}, \ldots[\text { Hint: Calculus required here. }]\) f) \(a_{0}^{2}, \quad 2 a_{0} a_{1}, \quad a_{1}^{2}+2 a_{0} a_{2}, \quad 2 a_{0} a_{3}+2 a_{1} a_{2}, \quad 2 a_{0} a_{4}+\) \(\quad 2 a_{1} a_{3}+a_{2}^{2}, \ldots\)
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