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Chapter 8: Problem 13

Use generating functions to determine the number of different ways 10 identical balloons can be given to four children if each child receives at least two balloons.

Short Answer

Expert verified
There are 10 ways to distribute the 10 balloons.

Step by step solution

01

Understanding the problem

Determine the number of ways to distribute 10 identical balloons to four children, each receiving at least 2 balloons.
02

Formulating the generating function

To use generating functions, first ensure each child receives at least 2 balloons. Represent the number of balloons a child receives as \(x_i\). Therefore, for each child \(x_i \geq 2 \). Define a new variable \(y_i = x_i - 2 \). This transformation changes the problem to finding non-negative integers for which the generating function is \( \sum_{x_i \geq 2} z^{x_i} \).
03

Adjusting the generating function

Since \(z^{x_i} = z^{2+y_i}\), the generating function simplifies to \[z^2 \sum_{y_i \geq 0} z^{y_i} = z^2 \left(\frac{1}{1-z}\right) = \frac{z^2}{1-z} \].
04

Creating the final generating function

Perform this for each child such that the generating function becomes \[ \left( \frac{z^2}{1-z} \right)^4 = \frac{z^8}{(1-z)^4} \]. We need the coefficient of \(z^{10}\).
05

Extracting the coefficient

The formula to extract the coefficient of \(z^{10}\) involves expanding \(\frac{z^8}{(1-z)^4} \). This can be written as \[ z^8 \left( \sum_{k=0}^{\infty} \binom{n+k-1}{k} z^k\right) \].
06

Identifying the necessary coefficient

Since we need the coefficient of \(z^{10}\) in the expanded form, identify the coefficient of \(z^2\) in \(\left( \sum_{k=0}^{\infty} \binom{n+k-1}{k} z^k\right) \) with \(n=4\). This requires us to find \(\binom{10-8+4-1}{10-8} = \binom{5}{2} = 10\).

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Key Concepts

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

Combinatorial Analysis
Combinatorial Analysis is a branch of mathematics dealing with counting, arrangement, and combination of objects. The goal is to determine the number of possible outcomes in various scenarios. When faced with problems like distributing balloons to children, combinatorial analysis helps to systematically break down the problem into manageable steps and find the solution. By using tools such as generating functions, we can tackle problems where traditional counting techniques may fall short. Generating functions transform the counting problem into a problem of finding coefficients in an algebraic expression, simplifying the process and uncovering patterns.
Balloon Distribution Problem
The balloon distribution problem is a common example in combinatorial analysis used to understand the distribution of identical objects (balloons) to distinct groups (children). Here, each child must receive at least a certain number of balloons. In our specific problem, each child must receive at least two balloons.

To tackle it, we redefined the problem to make it easier. By transforming the variables, we accounted for the minimum requirement directly in our generating function, simplifying the step-by-step calculation. This transformation lets us use a straightforward generating function to determine the number of acceptable distributions.
Coefficient Extraction
Coefficient extraction is the process of identifying specific coefficients from expanded algebraic expressions, particularly from generating functions. In our balloon distribution example, once we've set up the final generating function, we need to find the coefficient corresponding to the power representing the total number of balloons (in this case, 10 balloons).

The process involves working with series expansions and employing binomial coefficients to isolate the coefficient of interest. We look at the coefficient of the desired power in the final algebraic form to find our result. In mathematical terms, if our function is in the form \(g(z) = \sum_{k=0}^{\infty} a_k z^k\), extracting the coefficient of \(z^{10}\) means finding the value of \(a_{10}\).

For our example, we use the generating function \( \left( \frac{z^2}{1-z} \right)^4 = \frac{z^8}{(1-z)^4} \) and expand it to identify the coefficient of \(z^{10}\), which ultimately involves recognizing it as \( \binom{5}{2} = 10 \).

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Most popular questions from this chapter

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 if \(n\) is a positive integer, then the number of partitions of \(n\) into distinct parts equals the number of partitions of \(n\) into odd parts with repetitions allowed; that is, \(p_{o}(n)=p_{d}(n) .[\text { Hint: Show that the generating }\) functions for \(p_{o}(n)\) and \(p_{d}(n)\) are equal. \(]\)

This exercise deals with the problem of finding the largest sum of consecutive terms of a sequence of n real numbers. When all terms are positive, the sum of all terms provides the answer, but the situation is more complicated when some terms are negative. For example, the maximum sum of consecutive terms of the sequence \(-2,3,-1,6,-7,4\) is \(3+(-1)+6=8 .\) (This exercise is based on \([\mathrm{Be} 86] .\) Recall that in Exercise 56 in Section 8.1 we developed a dynamic programming algorithm for solving this problem. Here, we first look at the brute-force algorithm for solving this problem; then we develop a divide- and-conquer algorithm for solving it. a) Use pseudocode to describe an algorithm that solves this problem by finding the sums of consecutive terms starting with the first term, the sums of consecutive terms starting with the second term, and so on, keeping track of the maximum sum found so far as the algorithm proceeds. b) Determine the computational complexity of the algorithm in part (a) in terms of the number of sums computed and the number of comparisons made. c) Devise a divide-and-conquer algorithm to solve this problem. [Hint: Assume that there are an even number of terms in the sequence and split the sequence into two halves. Explain how to handle the case when the maximum sum of consecutive terms includes terms in both halves.] d) Use the algorithm from part (c) to find the maximum sum of consecutive terms of each of the sequences: \(-2,4,-1,3,5,-6,1,2 ; 4,1,-3,7,-1,-5, \quad 3,-2 ;\) and \(-1,6,3,-4,-5,8,-1,7\) e) Find a recurrence relation for the number of sums and comparisons used by the divide-and-conquer algorithm from part (c). f ) Use the master theorem to estimate the computational complexity of the divide-and-conquer algorithm. How does it compare in terms of computational complexity with the algorithm from part (a)?

(Calculus required) Let \(\left\\{C_{n}\right\\}\) be the sequence of Catalan numbers, that is, the solution to the recurrence relation \(C_{n}=\sum_{k=0}^{n-1} C_{k} C_{n-k-1}\) with \(C_{0}=C_{1}=1\) (see Example 5 in Section 8.1\()\) a) Show that if \(G(x)\) is the generating function for the sequence of Catalan numbers, then \(x G(x)^{2}-G(x)+\) \(1=0 .\) Conclude (using the initial conditions) that \(G(x)=(1-\sqrt{1-4 x}) /(2 x)\) b) Use Exercise 42 to conclude that $$ G(x)=\sum_{n=0}^{\infty} \frac{1}{n+1}\left(\begin{array}{c}{2 n} \\\ {n}\end{array}\right) x^{n} $$ so that $$ C_{n}=\frac{1}{n+1}\left(\begin{array}{c}{2 n} \\ {n}\end{array}\right) $$ c) Show that \(C_{n} \geq 2^{n-1}\) for all positive integers \(n\)

Use generating functions to find the number of ways to choose a dozen bagels from three varieties鈥攅gg, salty, and plain鈥攊f at least two bagels of each kind but no more than three salty bagels are chosen.

What is the generating function for the sequence \(\left\\{c_{k}\right\\}\) where \(c_{k}\) is the number of ways to make change for \(k\) dollars using \(\$ 1\) bills, \(\$ 2\) bills, \(\$ 5\) bills, and \(\$ 10\) bills?
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