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## 数学代写|组合学代写Combinatorics代考|Steiner Triple Systems

Let $Y$ be a finite set, and let $\mathcal{A}=\left(A_1, A_2, \ldots, A_n\right)$ be a family ${ }^6$ of $n$ subsets of $Y$. A family $\left(e_1, e_2, \ldots, e_n\right)$ of elements of $Y$ is called a system of representatives of $\mathcal{A}$, provided that
$e_1$ is in $A_1, e_2$ is in $A_2, \ldots, e_n$ is in $A_n$.
In a system of representatives, the element $e_i$ belongs to $A_i$ and thus “represents” the set $A_i$. If, in a system of representatives, the elements $e_1, e_2, \ldots, e_n$ are all different, then $\left(e_1, e_2, \ldots, e_n\right)$ is called a system of distinct representatives, abbreviated SDR.

Example. Let $\left(A_1, A_2, A_3, A_4\right)$ be the family of subsets of the set $Y={a, b, c, d, e}$, defined by
$$A_1={a, b, c}, A_2={b, d}, A_3={a, b, d}, A_4={b, d} .$$
Then $(a, b, b, d)$ is a system of representatives, and $(c, b, a, d)$ is an SDR.
A family $\mathcal{A}=\left(A_1, A_2, \ldots, A_n\right)$ of nonempty sets always has a system of representatives. We need pick only one element from each of the sets to obtain a system of representatives. However, the family $\mathcal{A}$ need not have an SDR even though all the sets in the family are nonempty. For instance, if there are two sets in the family, say, $A_1$ and $A_2$, each containing only one element, and the element in $A_1$ is the same as the element in $\mathrm{A}_2$, that is,
$$A_1={x}, A_2={x}$$
then the family $A$ does not have an SDR. This is because, in any system of representatives, $x$ has to represent both $A_1$ and $A_2$, and thus no SDR exists (no matter what $A_3, \ldots, A_n$ equal). However, a family $A$ can fail to have an SDR for somewhat more complicated reasons.

## 数学代写|组合学代写Combinatorics代考|Latin Squares

Latin squares were introduced in Section 1.5 in connection with Euler’s problem of the 36 officers, and the reader may wish to review that section before proceeding. A formal definition is the following: Let $n$ be a positive integer and let $S$ be a set of $n$ distinct elements. A Latin square of order $n$, based on the set $S$, is an $n$-by- $n$ array, each of whose entries is an element of $S$ such that each of the $n$ elements of $S$ occurs once (and hence exactly once) in each row and once in each column. Thus each of the rows and each of the columns of a Latin square is a permutation of the elements of $S$. It follows from the pigeonhole principle that we can check whether an $n$-by- $n$ array based on a set $S$ of $n$ elements is a Latin square in either of two ways: (i) check that each element of $S$ occurs at least once in each row and at least once in each column, or (ii) check that no element of $S$ occurs more than once in each row and no more than once in each column.

The actual nature of the elements of $S$ is of no importance and usually we take $S$ to be $Z_n={0,1, \ldots, n-1}$. In this case, we number the rows and the columns of the Latin square as $0,1, \ldots, n-1$, rather than the more conventional $1,2, \ldots, n$. A 1-by-1 array is always a Latin square based on the set consisting of its unique element. Other examples of Latin squares are the following:
$$\left[\begin{array}{ll} 0 & 1 \ 1 & 0 \end{array}\right],\left[\begin{array}{lll} 0 & 1 & 2 \ 1 & 2 & 0 \ 2 & 0 & 1 \end{array}\right],\left[\begin{array}{llll} 0 & 1 & 2 & 3 \ 1 & 2 & 3 & 0 \ 2 & 3 & 0 & 1 \ 3 & 0 & 1 & 2 \end{array}\right] .$$
To confirm our stated convention, row 0 of the last square is the permutation $0,1,2,3$, and row 2 is the permutation $2,3,0,1$.

## 数学代写|组合学代写Combinatorics代考|Latin Squares

$S$元素的实际性质并不重要，通常我们把$S$当作$Z_n={0,1, \ldots, n-1}$。在本例中，我们将拉丁方块的行和列编号为$0,1, \ldots, n-1$，而不是更传统的$1,2, \ldots, n$。1 × 1数组总是基于由其唯一元素组成的集合的拉丁正方形。其他关于拉丁方块的例子如下:
$$\left[\begin{array}{ll} 0 & 1 \ 1 & 0 \end{array}\right],\left[\begin{array}{lll} 0 & 1 & 2 \ 1 & 2 & 0 \ 2 & 0 & 1 \end{array}\right],\left[\begin{array}{llll} 0 & 1 & 2 & 3 \ 1 & 2 & 3 & 0 \ 2 & 3 & 0 & 1 \ 3 & 0 & 1 & 2 \end{array}\right] .$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|组合学代写Combinatorics代考|Systems of Distinct Representatives

Let $Y$ be a finite set, and let $\mathcal{A}=\left(A_1, A_2, \ldots, A_n\right)$ be a family ${ }^6$ of $n$ subsets of $Y$. A family $\left(e_1, e_2, \ldots, e_n\right)$ of elements of $Y$ is called a system of representatives of $\mathcal{A}$, provided that
$e_1$ is in $A_1, e_2$ is in $A_2, \ldots, e_n$ is in $A_n$.
In a system of representatives, the element $e_i$ belongs to $A_i$ and thus “represents” the set $A_i$. If, in a system of representatives, the elements $e_1, e_2, \ldots, e_n$ are all different, then $\left(e_1, e_2, \ldots, e_n\right)$ is called a system of distinct representatives, abbreviated SDR.

Example. Let $\left(A_1, A_2, A_3, A_4\right)$ be the family of subsets of the set $Y={a, b, c, d, e}$, defined by
$$A_1={a, b, c}, A_2={b, d}, A_3={a, b, d}, A_4={b, d} .$$
Then $(a, b, b, d)$ is a system of representatives, and $(c, b, a, d)$ is an SDR.
A family $\mathcal{A}=\left(A_1, A_2, \ldots, A_n\right)$ of nonempty sets always has a system of representatives. We need pick only one element from each of the sets to obtain a system of representatives. However, the family $\mathcal{A}$ need not have an SDR even though all the sets in the family are nonempty. For instance, if there are two sets in the family, say, $A_1$ and $A_2$, each containing only one element, and the element in $A_1$ is the same as the element in $\mathrm{A}_2$, that is,
$$A_1={x}, A_2={x}$$
then the family $A$ does not have an SDR. This is because, in any system of representatives, $x$ has to represent both $A_1$ and $A_2$, and thus no SDR exists (no matter what $A_3, \ldots, A_n$ equal). However, a family $A$ can fail to have an SDR for somewhat more complicated reasons.

## 数学代写|组合学代写Combinatorics代考|Stable Marriages

In this section ${ }^8$ we consider a variation of the marriage problem discussed in the previous section.

There are $n$ women and $n$ men in a community. Each woman ranks each man in accordance with her preference for that man as a spouse. No ties are allowed, so that if a woman is indifferent between two men, we nonetheless require that she express some preference. The preferences are to be purely ordinal, and thus each woman ranks the men in the order $1,2, \ldots, n$. Similarly, each man ranks the women in the order $1,2, \ldots, n$. There are $n$ ! ways in which the women and men can be paired so that a complete marriage takes place. We say that a complete marriage is unstable, provided that there exist two women $A$ and $B$ and two men $a$ and $b$ such that
(i) $A$ and $a$ get married;
(ii) $B$ and $b$ get married;
(iii) $A$ prefers (i.e., ranks higher) $b$ to $a$;
(iv) $b$ prefers $A$ to $B$.
Thus, in an unstable complete marriage, $A$ and $b$ could act independently of the others and run off with each other, since both would regard their new partner as more preferable than their current spouse. Thus, the complete marriage is “unstable” in the sense that it can be upset by a man and a woman acting together in a manner that is beneficial to both. A complete marriage is called stable, provided it is not unstable. The question that arises first is Does there always exist a stable, complete marriage?

We set up a mathematical model for this problem by using a bipartite graph again. Let $G=(X, \Delta, Y)$ be a bipartite graph in which
$$X=\left{w_1, w_2, \ldots, w_n\right}$$
is the set of $n$ women and
$$Y=\left{m_1, m_2, \ldots, m_n\right}$$is the set of $n$ men. We join each woman-vertex (left is now woman) to each man-vertex (right is now man). The resulting bipartite graph is complete in the sense that it contains all possible edges between its two sets of vertices. ${ }^9$ Corresponding to each edge $\left{w_i, m_j\right}$, there is a pair $p, q$ of numbers where $p$ denotes the position of $m_j$ in $w_i$ ‘s ranking of the men, and $q$ denotes the position of $w_i$ in $m_j$ ‘s ranking of the women. A complete marriage of the women and men corresponds to a perfect matching (of $n$ edges) in this bipartite graph $G$.

## 数学代写|组合学代写Combinatorics代考|Systems of Distinct Representatives

$e_1$在$A_1, e_2$在$A_2, \ldots, e_n$在$A_n$。

$$A_1={a, b, c}, A_2={b, d}, A_3={a, b, d}, A_4={b, d} .$$

$$A_1={x}, A_2={x}$$

## 数学代写|组合学代写Combinatorics代考|Stable Marriages

(一)$A$和$a$结婚;
(二)$B$和$b$结婚;
(iii)相对于$a$, $A$更倾向于(即排名更高)$b$;
(iv) $b$更喜欢$A$而不是$B$。

$$X=\left{w_1, w_2, \ldots, w_n\right}$$

$$Y=\left{m_1, m_2, \ldots, m_n\right}$$的集合是 $n$ 男人。我们将每个女人顶点(左边现在是女人)连接到每个男人顶点(右边现在是男人)。所得到的二部图是完备的，因为它包含了两组顶点之间所有可能的边。 ${ }^9$ 对应于每条边 $\left{w_i, m_j\right}$，有一对 $p, q$ 其中的数 $p$ 表示的位置 $m_j$ 在 $w_i$ 的排名，和 $q$ 表示的位置 $w_i$ 在 $m_j$ 这是对女性的排名。一个完整的男女婚姻对应于一个完美的匹配 $n$ 边)在这个二部图中 $G$．

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Combinatorics, 数学代写, 组合学

## avatest™帮您通过考试

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## 数学代写|组合学代写Combinatorics代考|Partition Numbers

A partition of a positive integer $n$ is a representation of $n$ as an unordered sum of one or more positive integers, called parts. Since the order of the parts is unimportant, we can always arrange the parts so that they are ordered from largest to smallest. The partitions of $1,2,3,4$, and 5 are, respectively, $1 ; 2,1+1 ; 3,2+1,1+1+1$; $4,3+1,2+2,2+1+1,1+1+1+1$; and $5,4+1,3+2,3+1+$ $1,2+2+1,2+1+1+1,1+1+1+1+1$.
A partition of $n$ is sometimes written as
$$\lambda=n^{a_n} \ldots 2^{a_2} \ldots 1^{a_1},$$
where $a_i$ is a nonnegative integer equal to the number of parts equal to $i$. (This expression is purely symbolic; its terms are not exponentials nor is the expression a product.) When written in the form (8.23), the term $i^{a_i}$ is usually omitted if $a_i=0$. In this notation, the partitions of 5 are:
$$5^1, 4^1 1^1, 3^1 2^1, 3^1 1^2, 2^2 1^1, 2^1 1^3, 1^5 .$$
Let $p_n$ denote the number of different partitions of the positive integer $n$, and for convenience, let $p_0=1$. The partition sequence is the sequence of numbers
$$p_0, p_1, \ldots, p_n, \ldots$$
By the preceding discussion, we have $p_0=1, p_1=1, p_2=2, p_3=$ $3, p_4=5$, and $p_5=7$. It is a simple observation (cf. (8.23)) that $p_n$ equals the number of solutions in nonnegative integers $a_n, \ldots, a_2, a_1$ of the equation
$$n a_n+\cdots 2 a_2+1 a_1=n .$$

## 数学代写|组合学代写Combinatorics代考|A Geometric Problem

In this section we shall obtain a combinatorial geometric interpretation of the sum
$$h_n^{(k)}=\left(\begin{array}{l} n \ 0 \end{array}\right)+\left(\begin{array}{l} n \ 1 \end{array}\right)+\cdots+\left(\begin{array}{l} n \ k \end{array}\right) \quad(0 \leq k \leq n)$$
of the first $k+1$ binomial coefficients with upper argument equal to $n$-that is, the sum of the first $k+1$ numbers in row $n$ of Pascal’s triangle. For each fixed $k$, we obtain a sequence
$$h_0^{(k)}, h_1^{(k)}, h_2^{(k)}, \ldots, h_n^{(k)}, \ldots$$
If $k=0$, we have
$$h_n^{(0)}=\left(\begin{array}{l} n \ 0 \end{array}\right)=1$$
and (8.25) is the sequence of all 1’s. If $k=1$, we obtain
$$h_n^{(1)}=\left(\begin{array}{l} n \ 0 \end{array}\right)+\left(\begin{array}{l} n \ 1 \end{array}\right)=n+1$$
If $k=2$, we have
\begin{aligned} h_n^{(2)} & =\left(\begin{array}{l} n \ 0 \end{array}\right)+\left(\begin{array}{l} n \ 1 \end{array}\right)+\left(\begin{array}{l} n \ 2 \end{array}\right) \ & =1+n+\frac{n(n-1)}{2} \ & =\frac{n^2+n+2}{2} . \end{aligned}
We also note that $h_0^{(k)}=1$ for all $k$. We use Pascal’s formula to determine the differences of (8.25):
\begin{aligned} \Delta h_n^{(k)} & =h_{n+1}^{(k)}-h_n^{(k)} \ & =\left(\begin{array}{c} n+1 \ 0 \end{array}\right)+\left(\begin{array}{c} n+1 \ 1 \end{array}\right)+\cdots+\left(\begin{array}{c} n+1 \ k \end{array}\right)-\left(\begin{array}{c} n \ 0 \end{array}\right)-\left(\begin{array}{c} n \ 1 \end{array}\right)-\cdots-\left(\begin{array}{c} n \ k \end{array}\right) \ & \left.=\left[\left(\begin{array}{c} n+1 \ 1 \end{array}\right)-\left(\begin{array}{c} n \ 1 \end{array}\right)\right]+\cdots+\left[\begin{array}{c} n+1 \ k \end{array}\right)-\left(\begin{array}{c} n \ k \end{array}\right)\right] \ & =\left(\begin{array}{c} n \ 0 \end{array}\right)+\cdots+\left(\begin{array}{c} n \ k-1 \end{array}\right) . \end{aligned}
Hence,
$$\Delta h_n^{(k)}=h_n^{(k-1)} \text {. }$$
It is a consequence of (8.26) that the difference table for the sequence
$$h_0^{(k)}, h_1^{(k)}, h_2^{(k)}, h_2^{(k)}, \ldots, h_n^{(k)}, \ldots$$
can be obtained from the difference table for
$$h_0^{(k-1)}, h_1^{(k-1)}, h_2^{(k-1)}, \ldots, h_n^{(k-1)}, \ldots$$
by inserting (8.27) on top as a new row.

## 数学代写|组合学代写Combinatorics代考|Partition Numbers

$$n a_n+\cdots 2 a_2+1 a_1=n .$$

## 数学代写|组合学代写Combinatorics代考|A Geometric Problem

$$h_n^{(k)}=\left(\begin{array}{l} n \ 0 \end{array}\right)+\left(\begin{array}{l} n \ 1 \end{array}\right)+\cdots+\left(\begin{array}{l} n \ k \end{array}\right) \quad(0 \leq k \leq n)$$

$$h_0^{(k)}, h_1^{(k)}, h_2^{(k)}, \ldots, h_n^{(k)}, \ldots$$

$$h_n^{(0)}=\left(\begin{array}{l} n \ 0 \end{array}\right)=1$$

$$h_n^{(1)}=\left(\begin{array}{l} n \ 0 \end{array}\right)+\left(\begin{array}{l} n \ 1 \end{array}\right)=n+1$$

\begin{aligned} h_n^{(2)} & =\left(\begin{array}{l} n \ 0 \end{array}\right)+\left(\begin{array}{l} n \ 1 \end{array}\right)+\left(\begin{array}{l} n \ 2 \end{array}\right) \ & =1+n+\frac{n(n-1)}{2} \ & =\frac{n^2+n+2}{2} . \end{aligned}

\begin{aligned} \Delta h_n^{(k)} & =h_{n+1}^{(k)}-h_n^{(k)} \ & =\left(\begin{array}{c} n+1 \ 0 \end{array}\right)+\left(\begin{array}{c} n+1 \ 1 \end{array}\right)+\cdots+\left(\begin{array}{c} n+1 \ k \end{array}\right)-\left(\begin{array}{c} n \ 0 \end{array}\right)-\left(\begin{array}{c} n \ 1 \end{array}\right)-\cdots-\left(\begin{array}{c} n \ k \end{array}\right) \ & \left.=\left[\left(\begin{array}{c} n+1 \ 1 \end{array}\right)-\left(\begin{array}{c} n \ 1 \end{array}\right)\right]+\cdots+\left[\begin{array}{c} n+1 \ k \end{array}\right)-\left(\begin{array}{c} n \ k \end{array}\right)\right] \ & =\left(\begin{array}{c} n \ 0 \end{array}\right)+\cdots+\left(\begin{array}{c} n \ k-1 \end{array}\right) . \end{aligned}

$$\Delta h_n^{(k)}=h_n^{(k-1)} \text {. }$$

$$h_0^{(k)}, h_1^{(k)}, h_2^{(k)}, h_2^{(k)}, \ldots, h_n^{(k)}, \ldots$$

$$h_0^{(k-1)}, h_1^{(k-1)}, h_2^{(k-1)}, \ldots, h_n^{(k-1)}, \ldots$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|组合学代写Combinatorics代考|Solving a Single Recursion

In this section we’ll use ordinary generating functions to solve some simple recursions, including two that we were unable to solve previously: the Fibonacci numbers and the number of unlabeled full binary RP-trees.

Example 10.2 Fibonacci numbers Let $F_n$ be the number of $n$ long sequences of zeroes and ones with no consecutive ones. We can easily see that $F_1=2$ and $F_2=3$, but what is the general formula?

Suppose that $t_1, \ldots, t_n$ is an arbitrary sequence of desired form. We want to see what happens when we remove the end of the sequence, so we assume that $n>1$. If $t_n=0$, then $t_1, \ldots, t_{n-1}$ is also an arbitrary sequence of the desired form. Now suppose that $t_n=1$. Then $t_{n-1}=0$ and so, if $n>2, t_1, \ldots, t_{n-2}$ is an arbitrary sequence of the desired form. All this is reversible: Suppose that $n>2$. The following two operations produce all $n$ long sequences of the desired form exactly once.

• Let $t_1, \ldots, t_{n-1}$ be an arbitrary sequence of the desired form. Set $t_n=0$.
• Let $t_1, \ldots, t_{n-2}$ be an arbitrary sequence of the desired form. Set $t_{n-1}=0$ and $t_n=1$.
Since all $n$ long sequences of the desired form are obtained exactly once this way, the Rule of Sum yields the recursion
$$F_n=F_{n-1}+F_{n-2} \text { for } n>2$$

## 数学代写|组合学代写Combinatorics代考|Manipulating Generating Functions

Suppose we have an equation that determines a generating function $B(x)$; for example, $B(x)=\frac{1-\sqrt{1-4 x}}{2}$. The basic idea for obtaining a recursion for $B(x)$ is to rewrite the equation so that $B(x)$ appears in expressions that are simple and so that the remaining expressions are easy to expand in power series. Once a simple form has been found, equate coefficients of $x^n$ on the two sides of the equation. We’ll explore this idea here.

Example 10.6 Rational functions and recursions Suppose that $B(x)=P(x) / Q(x)$ where $P(x)$ and $Q(x)$ are polynomials. Expressions that involve division are usually not easy to expand unless the divisor is a product of linear factors with integer coefficients. Thus, we would usually rewrite our equation as $Q(x) B(x)=P(x)$ and then equate coefficients. This gives us a recursion for the $b_i$ ‘s which is linear and has constant coefficients.

The description of the procedure is a bit vague, so let’s look at an example. When we study systems of recursions in the next chapter, we will show that the number of ways to place nonoverlapping dominoes on a 2 by $n$ board has the generating function
$$C(x)=\frac{1-x}{1-3 x-x^2+x^3} .$$
Thus $P(x)=1-x$ and $Q(x)=1-3 x-x^2+x^3$. Using our plan, we have
$$\left(1-3 x-x^2+x^3\right) C(x)=1-x .$$
There are now various ways we can proceed:
Keep all subscripts nonnegative: When $n \geq 3$, the coefficient of $x^n$ on the right side is 0 and the coefficient on the left side is $c_n-3 c_{n-1}-c_{n-2}+c_{n-3}$, so all the subscripts are nonnegative. Rearranging this,
$$c_n=3 c_{n-1}+c_{n-2}-c_{n-3} \text { for } n \geq 3$$
The values of $a_0, a_1$ and $a_2$ are given by initial conditions. Looking at the coefficients of $x^0, x^1$ and $x^2$ on both sides of $(10.19)$, we have
$$a_0=1 \quad a_1-3 a_0=-1 \quad a_2-3 a_1-a_0=0$$

## 数学代写|组合学代写Combinatorics代考|Solving a Single Recursion

$$F_n=F_{n-1}+F_{n-2} \text { for } n>2$$

## 数学代写|组合学代写Combinatorics代考|Manipulating Generating Functions

$$C(x)=\frac{1-x}{1-3 x-x^2+x^3} .$$

$$\left(1-3 x-x^2+x^3\right) C(x)=1-x .$$

$$c_n=3 c_{n-1}+c_{n-2}-c_{n-3} \text { for } n \geq 3$$
$a_0, a_1$和$a_2$的值由初始条件给出。看看$(10.19)$两边的$x^0, x^1$和$x^2$的系数，我们有
$$a_0=1 \quad a_1-3 a_0=-1 \quad a_2-3 a_1-a_0=0$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|组合学代写Combinatorics代考|Proving That a Network Sorts

Unlike many software sorts, it is frequently difficult to prove that a network actually sorts all inputs correctly. There is no panacea for this problem. The following two theorems are sometimes helpful.
Theorem 8.2 Zero-One Principle ones, then it correctly sorts all inputs.
If a network correctly sorts all inputs of zeroes and
Theorem 8.3 Adjacent Comparisons If the comparators in a network only connect adjacent lines and if the network correctly sorts the reversed sequence $n, \ldots, 2,1$, then it correctly sorts all inputs.

We will prove the Zero-One Principle shortly. The proof of the other theorem is more complicated and will not be given.

Since the Adjacent Comparisons Theorem requires that only one input be checked, it is quite useful. Unfortunately, the comparators must connect adjacent lines. To see that this is needed, consider a three input network in which the top and bottom lines are connected by a comparator and there are no other comparators. It correctly sorts the inputs $1,2,3$ and $3,2,1$, but it does not sort any other permutations of $1,2,3$ correctly.

The Zero-One Principle may seem somewhat useless because it still requires that many inputs be considered. We will see that it is quite useful for proving that a Batcher sort works.

Proof: We now prove the Zero-One Principle. If the network fails to sort some sequence, we will show how to construct a sequence of zeroes and ones that it fails to sort.

The idea behind our proof is the following simple observation: Suppose that $f$ is a nondecreasing function, then a comparator treats $f(s)$ and $f(t)$ the same as it does $s$ and $t$. This is illustrated in Figure 8.5. It is easy to show by considering the three cases $st$.

Suppose a network has $N$ comparators, has inputs $x_1, \ldots, x_n$ and outputs $y_1, \ldots, y_n$. Let $f$ be a nondecreasing function. We will use induction on $N$ to prove that if $f\left(x_1\right), \ldots, f\left(x_n\right)$ are fed into the network, then $f\left(y_1\right), \ldots, f\left(y_n\right)$ emerge at the other end.

## 数学代写|组合学代写Combinatorics代考|Traversing Trees

A tree traversal algorithm is a systematic method for visiting all the vertices in an RP-tree. We’ve already seen a nonrecursive traversal algorithm in Theorem 3.5 (p.85). As we shall soon see a recursive description is much simpler. It is essentially a local description.

Traversal algorithms fall into two categories called “breadth first” and “depth first,” with depth first being the more common type. After explaining the categories, we’ll focus on depth first algorithms.

The left side of Figure 9.2 shows an RP-tree. Consider the right side of Figure 9.2. There we see the same RP-tree. Take your pencil and, starting at the root $a$, follow the arrows in such a way that you visit the vertices in the order
$$\text { abe } b f j f k f l f b a c a d g d h d i d a \text {. }$$
This manner of traversing the tree diagram, which extends in an obvious manner to any RP-tree $T$, is called a depth-first traversal of the ordered rooted tree $T$. The sequence of vertices (9.1) associated with the depth-first traversal of the RP-tree $T$ will be called the depth-first vertex sequence of $T$ and will be denoted by DFV $(T)$. If you do a depth-first traversal of an RP-tree $T$ and list the edges encountered (list an edge each time your pencil passes its midpoint in the diagram), you obtain the depth-first edge sequence of $T$, denoted by $\operatorname{DFE}(T)$. In Figure 9.2 , the sequence $\mathrm{DFE}(T)$ is
$$\begin{array}{lllllllllll} {a, b} & {b, e} & {b, e} & {b, f} & {f, j} & {f, j} & {f, k} & {f, k} & {f, l} & {f, l} & {b, f} \ {a, b} & {a, c} & {a, c} & {a, d} & {d, g} & {d, g} & {d, h} & {d, h} & {d, i} & {d, i} & {a, d} . \end{array}$$

The other important linear order associated with RP-trees is called breadth-first order. This order is obtained, in the case of Figure 9.2 , by reading the vertices or edges level by level, starting with the root. In the case of vertices, we obtain the breadth-first vertex sequence $(\mathrm{BFV}(T))$. In Figure 9.2 , $\mathrm{BFV}(T)=$ abcdefghijkl. Similarly, we can define the breadth-first edge sequence $(\mathrm{BFE}(T))$.

Although we have defined these orders for trees, the ideas can be extended to other graphs. For example, one can use a breadth first search to find the shortest (least number of choices) route out of a maze: Construct a decision tree in which each vertex corresponds to an intersection in the maze. (More than one vertex may correspond to the same intersection.) A vertex corresponding to an intersection already encountered in the breadth first search has no sons. The decisions at an intersection not previously encountered are all possibilities of the form “follow a passage to the next intersection.”

## 数学代写|组合学代写Combinatorics代考|Proving That a Network Sorts

0 – 1原则可能看起来有些无用，因为它仍然需要考虑许多输入。我们将看到，这对于证明批处理排序是非常有用的。

## 数学代写|组合学代写Combinatorics代考|Traversing Trees

$$\text { abe } b f j f k f l f b a c a d g d h d i d a \text {. }$$

$$\begin{array}{lllllllllll} {a, b} & {b, e} & {b, e} & {b, f} & {f, j} & {f, j} & {f, k} & {f, k} & {f, l} & {f, l} & {b, f} \ {a, b} & {a, c} & {a, c} & {a, d} & {d, g} & {d, g} & {d, h} & {d, h} & {d, i} & {d, i} & {a, d} . \end{array}$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|组合学代写Combinatorics代考|Motivation and Proof of the Theorem

Our proof of Theorem 8.1 will be by induction. Induction requires knowing the result (in this case
AC(n) ≥ log2 (n!)) beforehand. How would one ever come up with the result beforehand? A result like the Four Color Theorem (p. 158) might be conjectured after some experimentation, but one is unlikely to stumble upon Theorem 8.1 experimentally. We will “discover” it by relating sorting algorithms to decision trees, which lead easily to the inequality WC(n) ≥ log2 (n!). One might then test AC(n) ≥ log2 (n!) for small values of n, thus motivating Theorem 8.1. Someone versed in information theory might motivate the theorem as follows: “A comparison gives us one bit of information. Given k bits of information, we can distinguish among at most 2k different things. Since we must distinguish among n! different arrangements, we require that 2k ≥ n! and so k ≥ log2 (n!).”
(This is motivation, not a proof—it’s not even clear if it’s referring to worst case or average case behavior.) Let’s get on with things.
Suppose we are given a comparison based sorting algorithm. Since it is assumed to correctly sort n-lists, it must correctly sort lists in which all items are different. By simply renaming our items 1, 2, . . ., n, we can suppose that we are sorting lists which are permutations of n.
Our proof will make use of the decision tree associated with this sorting algorithm. We construct the tree as follows. Whenever we make a comparison in the course of the algorithm, our subsequent action depends on whether an inequality holds or fails to hold. Thus there are two possible decisions at each comparison and so each vertex in our decision tree has at most two sons.
Label each leaf of the tree with the permutations that led to that leaf and throw away any leaves that are not associated with any permutation. To do this, we start at the root with a permutation f of n and at each vertex in the tree we go left if the inequality we are checking holds and go right if it fails to hold. At the same time, we carry out whatever manipulations on the data in f that the algorithm requires. When we arrive at a leaf, the data in f will be sorted. Label the leaf with the f we started out with at the root, written in one line form. Do this for all n! permutations of n. For example, consider the following algorithm for sorting a permutation of 3.

1. If the entry in the first position exceeds the entry in the third position, switch them.
2. If the entry in the first position exceeds the entry in the second position, switch them.
3. If the entry in the second position exceeds the entry in the third position, switch them.

## 数学代写|组合学代写Combinatorics代考|Software Sorts

A merge sort algorithm was studied in Example 7.13 (p. 211). You should review it now.
All reasonably fast software sorts use a divide and conquer method for attacking the problem. As you may recall, divide and conquer means splitting the problem into a few smaller problems which are easier either because they are smaller or because they are simpler. In problems where divide and conquer is most successful, it is often the case that the smaller problems are simply instances of the same type of problem and they are handled by applying the algorithm recursively. To give you a bit better idea of what divide and conquer means, here is how the algorithms we’ll discuss use it. Some of this may not mean much to you until you’ve finished this section, so you may want to reread this list later. This is by no means an exhaustive list of the different types of software sorts.

Quicksort and merge sorts split the data and spend most of their time sorting the separate pieces. Thus they divide and conquer by producing two smaller sorting problems which are handled in a recursive manner.
Quicksort spends a little time dividing the data in such a way that recombining the two pieces after they are sorted is immediate. It divides the items into two collections so that all the items in the first collection should precede all the items in the second. The division is done “in place” by interchanging items that are in the wrong lists. Unless it is extremely unlucky, the two collections will have roughly the same number of elements. The two collections are then sorted separately.
Merge sorts reverse this: dividing is immediate and recombination takes a little time. In both cases, the “little time” required is proportional to the number of items being sorted because it requires a number of comparisons that is nearly equal to the number of items being sorted.

• An insertion sort builds up the sorted list by taking the items on the unsorted list one at a time and inserting them in a sorted list it is building. Divide and conquer can be used in the insertion process: To do a binary insertion sort, split the list into two nearly equal parts, decide which sublist should contain the new item, and iterate the process, using the sublist as the list.
• Suppose we are sorting a list of words (or numbers). Bucket sort focuses on one position in the words at a time. This is not usually a good divide and conquer approach because the task is not divided into just a few subproblems of roughly equal difficulty: On an n-long list with k characters per word, we focus in turn on each of the k positions. When n is large, k will be large, too.
It is easy to get a time estimate for the algorithm. The amount of time it takes to process one character position for all n words is proportional to n. Thus, the time to sort is proportional to nk. How fast a bucket sort is depends on how large k is compared to n.
• First, the items are arranged in a structure, called a “heap,” which is a rooted tree such that the smallest item in the tree is at the root and each of the sons of the root is also the root of a heap.
• Second, the items are removed from the heap one by one, starting with the top and preserving the heap structure.

## 数学代写|组合学代写Combinatorics代考|Motivation and Proof of the Theorem

AC(n)≥log2 (n!))一个人怎么能事先得出结果呢?像四色定理(第158页)这样的结果可以在一些实验之后推测出来，但是人们不可能在实验中偶然发现定理8.1。我们将通过将排序算法与决策树联系起来来“发现”它，这很容易导致不等式WC(n)≥log2 (n!)。对于n的小值，可以检验AC(n)≥log2 (n!)，从而推导出定理8.1。精通信息论的人可能会这样解释这个定理:“比较给了我们一点信息。给定k位信息，我们最多可以区分2k种不同的事物。因为我们必须区分n!不同的排列方式，我们要求2k≥n!所以k≥log2 (n!)
(这是动机，而不是证据——甚至不清楚它指的是最坏情况还是一般情况下的行为。)让我们开始工作吧。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|组合学代写Combinatorics代考|Some Number Sequences

Let
$$h_0, h_1, h_2, \ldots, h_n, \ldots$$
denote a sequence of numbers. We call $h_n$ the general term or generic term of the sequence. Two familiar types of sequences are
arithmetic sequences, in which each term is a constant $q$ more than the previous term,
and
geometric sequences, in which each term is a constant multiple $q$ of the previous term.
In both instances, the sequence is uniquely determined once the initial term $h_0$ and the constant $q$ are specified:
(arithmetic sequence)
$$h_0, h_0+q, h_0+2 q, \ldots, h_0+n q, \ldots$$
(geometric sequence)
$$h_0, q h_0, q^2 h_0, \ldots, q^n h_0, \ldots$$
In the case of an arithmetic sequence, we have the rule
$$h_n=h_{n-1}+q, \quad(n \geq 1)$$
and the general term is
$$h_n=h_0+n q, \quad(n \geq 0) .$$
In the case of a geometric sequence, we have the rule
$$h_n=q h_{n-1}, \quad(n \geq 1)$$
and the general term is
$$h_n=h_0 q^n, \quad(n \geq 0)$$

## 数学代写|组合学代写Combinatorics代考|Linear Homogeneous Recurrence Relations

Let
$$h_0, h_1, h_2, \ldots, h_n, \ldots$$
be a sequence of numbers. This sequence is said to satisfy a linear recurrence relation of order $k$, provided that there exist quantities $a_1, a_2, \ldots, a_k$, with $a_k \neq 0$, and a quantity $b_n$ (each of these quantities $a_1, a_2, \ldots, a_k, b_n$ may depend on $n$ ) such that
$$h_n=a_1 h_{n-1}+a_2 h_{n-2}+\cdots+a_k h_{n-k}+b_n, \quad(n \geq k) .$$
Example. The sequence of derangement numbers
$$D_0, D_1, D_2, \ldots, D_n, \ldots$$
satisfies the two recurrence relations
\begin{aligned} & D_n=(n-1) D_{n-1}+(n-1) D_{n-2}, \quad(n \geq 2) \ & D_n=n D_{n-1}+(-1)^n, \quad(n \geq 1) \text {. } \ & \end{aligned}
The first recurrence relation has order 2 and we have $a_1=n-1$, $a_2=n-1$ and $b_n=0$. The second recurrence relation has order 1 and we have $a_1=n$ and $b_n=(-1)^n$.
Example. The Fibonacci sequence $f_0, f_1, f_2, \ldots, f_n, \ldots$ satisfies the recurrence relation
$$f_n=f_{n-1}+f_{n-2} \quad(n \geq 2)$$
of order 2 with $a_1=1, a_2=1$, and $b_n=0$.

## 数学代写|组合学代写Combinatorics代考|Some Number Sequences

$$h_0, h_1, h_2, \ldots, h_n, \ldots$$

(等差数列)
$$h_0, h_0+q, h_0+2 q, \ldots, h_0+n q, \ldots$$
(几何序列)
$$h_0, q h_0, q^2 h_0, \ldots, q^n h_0, \ldots$$

$$h_n=h_{n-1}+q, \quad(n \geq 1)$$

$$h_n=h_0+n q, \quad(n \geq 0) .$$

$$h_n=q h_{n-1}, \quad(n \geq 1)$$

$$h_n=h_0 q^n, \quad(n \geq 0)$$

## 数学代写|组合学代写Combinatorics代考|Linear Homogeneous Recurrence Relations

$$h_0, h_1, h_2, \ldots, h_n, \ldots$$

$$h_n=a_1 h_{n-1}+a_2 h_{n-2}+\cdots+a_k h_{n-k}+b_n, \quad(n \geq k) .$$

$$D_0, D_1, D_2, \ldots, D_n, \ldots$$

\begin{aligned} & D_n=(n-1) D_{n-1}+(n-1) D_{n-2}, \quad(n \geq 2) \ & D_n=n D_{n-1}+(-1)^n, \quad(n \geq 1) \text {. } \ & \end{aligned}

$$f_n=f_{n-1}+f_{n-2} \quad(n \geq 2)$$
2阶的 $a_1=1, a_2=1$，和 $b_n=0$．

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|组合学代写Combinatorics代考|The Inclusion-Exclusion Principle

Alternatively, we could make an indirect count by observing that the number of permutations of ${1,2, \ldots, n}$ with 1 in the first position is the same as the number $(n-1)$ ! of permutations of ${2,3, \ldots, n}$. Since the total number of permutations of ${1,2, \ldots, n}$ is $n !$, the number of permutations of ${1,2, \ldots, n}$ in which 1 is not in the first position is $n !-(n-1) !=(n-1) !(n-1)$.

Example. Count the number of integers between 1 and 600 , inclusive, which are not divisible by 6 .

We can do this indirectly as follows. The number of integers between 1 and 600 which are divisible by 6 is $600 / 6=100$ since every sixth integer is divisible by 6 . Hence $600-100=500$ of the integers between 1 and 600 are not divisible by 6 .

The rule used to obtain an indirect count in these examples is the following. If $A$ is a subset of a set $S$, then the number of objects in $A$ equals the number of objects in $S$ minus the number not in $A$. Recall that
$$\bar{A}=S-A={x: x \text { in } S \text { but } x \text { not in } A}$$
is the complement of $A$ in $S$-that is, the set consisting of those objects in $S$ which are not in $A$. The rule can then be written as
$$|A|=|S|-|\bar{A}| \text { or, equivalently, }|\bar{A}|=|S|-|A| \text {. }$$
This formula is the simplest instance of the inclusion-exclusion principle.

## 数学代写|组合学代写Combinatorics代考|Combinations with Repetition

In Sections 3.3 and 3.5 we have shown that the number of $r$-combinations of a set of $n$ distinct elements is
$$\left(\begin{array}{l} n \ r \end{array}\right)=\frac{n !}{r !(n-r) !}$$
and that the number of $r$-combinations of a multiset with $k$ distinct objects, each with an infinite repetition number, equals
$$\left(\begin{array}{c} r+k-1 \ r \end{array}\right) \text {. }$$
In this section we show how the latter formula, in conjunction with the inclusion-exclusion principle, gives a method for finding the number of $r$-combinations of a multiset without any restrictions on its repetition numbers.

Suppose $T$ is a multiset and an object $x$ of $T$ of a certain type has a repetition number that is greater than $r$. Then the number of $r$ combinations of $T$ equals the number of $r$-combinations of the multiset obtained from $T$ by replacing the repetition number of $x$ by $r$. This is so because the number of times $x$ can be used in an $r$-combination of $T$ cannot exceed $r$. Therefore, any repetition number that is greater than $r$ can be replaced by $r$. For example, the number of 8-combinations of the multiset ${3 \cdot a, \infty \cdot b, 6 \cdot c, 10 \cdot d, \infty \cdot e}$ is the same as the number of $8-$ combinations of the multiset ${3 \cdot a, 8 \cdot b, 6 \cdot c, 8 \cdot d, 8 \cdot e}$. We can summarize by saying that we have determined the number of $r$-combinations of a multiset $T=\left{n_1 \cdot a_1, n_2 \cdot a_2, \ldots, n_k \cdot a_k\right}$ in the two “extreme” cases:
(i) $n_1=n_2=\cdots=n_k=1$; (i.e., $T$ is a set) and
(ii) $n_1=n_2=\cdots=n_k=r$.

## 数学代写|组合学代写Combinatorics代考|The Inclusion-Exclusion Principle

$$\bar{A}=S-A={x: x \text { in } S \text { but } x \text { not in } A}$$

$$|A|=|S|-|\bar{A}| \text { or, equivalently, }|\bar{A}|=|S|-|A| \text {. }$$

## 数学代写|组合学代写Combinatorics代考|Combinations with Repetition

$$\left(\begin{array}{l} n \ r \end{array}\right)=\frac{n !}{r !(n-r) !}$$

$$\left(\begin{array}{c} r+k-1 \ r \end{array}\right) \text {. }$$

(i) $n_1=n_2=\cdots=n_k=1$;(例如，$T$是一个集合)和
(ii) $n_1=n_2=\cdots=n_k=r$。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|组合学代写Combinatorics代考|Partial Orders and Equivalence Relations

In this chapter we have defined various “natural” orders on the sets of permutations, combinations, and $r$-combinations of a finite set, namely, the orders determined by the generating schemes. These orders are “total orders” in the sense that there is a first object, a second object, a third object, …, a last object. There is a more general notion of order, called “partial order,” which is extremely important and useful in mathematics. Perhaps the two partial orders which are not total orders that are most familiar are those defined by containment of one set in another and divisibility of one integer by another. These are partial orders in the sense that, given any two sets, neither may be a subset of the other, and given any two integers, neither may be divisible by the other.

In order to give a precise definition of a partial order, it is important to know what is meant in mathematics by a relation. Let $X$ be a set. A relation $R$ on $X$ is a “property” that may or may not hold between any two given elements of $X$. More formally, a relation on $X$ is a subset $R$ of the set $X \times X$ of ordered pairs of elements of $X$. We write $a R b$, provided that the ordered pair $(a, b)$ belongs to $R$; we also write $a \not R b$ whenever $(a, b)$ is not in $R$.

Example. Let $X={1,2,3,4,5,6}$. Write $a \mid b$ to mean that $a$ is a divisor of $b$ (equivalently, $b$ is divisible by $a$ ). This defines a partial order on $X$ and we have, for example, $2 \mid 6$ and $3 \not 5$.

Now consider the collection $\mathcal{P}(X)$ of all subsets (i.e., combinations) of $X$. For $A$ and $B$ in $\mathcal{P}(X)$, we write as usual $A \subseteq B$, read $A$ is contained in $B$, provided that every element of $A$ is also an element of $B$. This defines a relation on $\mathcal{P}(X)$ and we have, for example, ${1} \subseteq{1,3}$ and ${1,2} \nsubseteq{2,3}$.

The following are special properties that a relation $R$ on a set $X$ may have:

1. $R$ is reflexive, provided that $x R x$ for all $x$ in $X$.
2. $R$ is irreflexive, provided that $x \not R x$ for all $x$ in $X$.
3. $R$ is symmetric, provided that, for all $x$ and $y$ in $X$, whenever we have $x R y$ we also have $y R x$.
4. $R$ is antisymmetric, provided that, for all $x$ and $y$ in $X$ with $x \neq y$, whenever we have $x R y$, we also have $y \not R x$. Equivalently, for all $x$ and $y$ in $X, x R y$ and $y R x$ together imply that $x=y$.
5. $R$ is transitive, provided that, for all $x, y, z$ in $X$, whenever we have $x R y$ and $y R z$, we also have $x R z$.

## 数学代写|组合学代写Combinatorics代考|Pascal’s Formula

The binomial coefficients $\left(\begin{array}{l}n \ k\end{array}\right)$ have been defined in Section 3.3 for all nonnegative integers $k$ and $n$. Recall that $\left(\begin{array}{l}n \ k\end{array}\right)=0$ if $k>n$ and that $\left(\begin{array}{l}n \ 0\end{array}\right)=1$ for all $n$. If $n$ is positive and $1 \leq k \leq n$, then
$$\left(\begin{array}{l} n \ k \end{array}\right)=\frac{n !}{k !(n-k) !}=\frac{n(n-1) \cdots(n-k+1)}{k(k-1) \cdots 1} .$$
In Section 3.3, we noted that
$$\left(\begin{array}{l} n \ k \end{array}\right)=\left(\begin{array}{c} n \ n-k \end{array}\right)$$
This relation is valid for all integers $k$ and $n$ with $0 \leq k \leq n$.

Theorem 5.1.1 (Pascal’s formula) For all integers $n$ and $k$ with $1 \leq k \leq n-1$,
$$\left(\begin{array}{l} n \ k \end{array}\right)=\left(\begin{array}{c} n-1 \ k \end{array}\right)+\left(\begin{array}{l} n-1 \ k-1 \end{array}\right) .$$
Proof. One way to prove this identity is to substitute the values of the binomial coefficients and then check that both sides are equal. We leave this straightforward verification to the reader.

A combinatorial proof can be obtained as follows: Let $S$ be a set of $n$ elements. We distinguish one of the elements of $S$ and denote it by $x$. We then partition the set $X$ of $k$-combinations of $S$ into two parts, $A$ and $B$. In $A$ we put all those $k$-combinations which do not contain $x$. In $B$ we put all the $k$-combinations which do contain $x$. The size of $X$ is $|X|=\left(\begin{array}{l}n \ k\end{array}\right)$; hence, by the addition principle,
$$\left(\begin{array}{l} n \ k \end{array}\right)=|A|+|B|$$
The $k$-combinations in $A$ are exactly the $k$-combinations of the set $S-{x}$ of $n-1$ elements; thus, the size of $A$ is
$$|A|=\left(\begin{array}{c} n-1 \ k \end{array}\right)$$
A $k$-combination in $B$ is obtained by adjoining the element $x$ to a $(k-1)$-combination of $S-{x}$. Hence, the size of $B$ satisfies
$$|B|=\left(\begin{array}{l} n-1 \ k-1 \end{array}\right)$$
Combining these facts, we obtain
$$\left(\begin{array}{l} n \ k \end{array}\right)=\left(\begin{array}{c} n-1 \ k \end{array}\right)+\left(\begin{array}{l} n-1 \ k-1 \end{array}\right)$$

## 数学代写|组合学代写Combinatorics代考|Partial Orders and Equivalence Relations

$R$ 是自反的，前提是$x R x$适用于$X$中的所有$x$。

$R$ 是不自反的，前提是$x \not R x$对于$X$中的所有$x$。

$R$ 是对称的，只要对于$X$中的所有$x$和$y$，只要我们有$x R y$，我们就有$y R x$。

$R$ 是反对称的，只要，对于$X$和$x \neq y$中的所有$x$和$y$，只要我们有$x R y$，我们也有$y \not R x$。同样，对于$X, x R y$和$y R x$中的所有$x$和$y$，一起意味着$x=y$。

$R$ 是可传递的，前提是，对于$X$中的所有$x, y, z$，只要我们有$x R y$和$y R z$，我们也有$x R z$。

## 数学代写|组合学代写Combinatorics代考|Pascal’s Formula

$$\left(\begin{array}{l} n \ k \end{array}\right)=\frac{n !}{k !(n-k) !}=\frac{n(n-1) \cdots(n-k+1)}{k(k-1) \cdots 1} .$$

$$\left(\begin{array}{l} n \ k \end{array}\right)=\left(\begin{array}{c} n \ n-k \end{array}\right)$$

$$\left(\begin{array}{l} n \ k \end{array}\right)=\left(\begin{array}{c} n-1 \ k \end{array}\right)+\left(\begin{array}{l} n-1 \ k-1 \end{array}\right) .$$

$$\left(\begin{array}{l} n \ k \end{array}\right)=|A|+|B|$$
$A$中的$k$ -组合就是$n-1$元素集合$S-{x}$的$k$ -组合;因此，$A$的大小为
$$|A|=\left(\begin{array}{c} n-1 \ k \end{array}\right)$$

$$|B|=\left(\begin{array}{l} n-1 \ k-1 \end{array}\right)$$

$$\left(\begin{array}{l} n \ k \end{array}\right)=\left(\begin{array}{c} n-1 \ k \end{array}\right)+\left(\begin{array}{l} n-1 \ k-1 \end{array}\right)$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|组合学代写Combinatorics代考|Permutations of Sets

Let $r$ be a positive integer. By an $r$-permutation of a set $S$ of $n$ elements we understand an ordered arrangement of $r$ of the $n$ elements. If $S={a, b, c}$, then the three 1-permutations of $S$ are
$a b c$,
the six 2-permutations of $S$ are
$a b \quad a c \quad b a \quad b c \quad c a \quad c b$,
and the six 3-permutations of $S$ are
$a b c \quad a c b \quad b a c$ bca $c a b \quad c b a$.
There are no 4-permutations of $S$ since $S$ has fewer than 4 elements.
We denote by $P(n, r)$ the number of $r$-permutations of an $n$-element set. If $r>n$, then $P(n, r)=0$. Clearly $P(n, 1)=n$ for each positive integer $n$. An $n$-permutation of an $n$-element set $S$ will be more simply called a permutation of $S$ or a permutation of $n$ elements. Thus, a permutation of a set $S$ can be thought of as a listing of the elements of $S$ in some order. Previously we saw that $P(3,1)=3, P(3,2)=6$, and $P(3,3)=6$.
Theorem 3.2.1 For $n$ and $r$ positive integers with $r \leq n$,
$$P(n, r)=n \times(n-1) \times \cdots \times(n-r+1) .$$
Proof. In constructing an $r$-permutation of an $n$-element set, we can choose the first item in $n$ ways, the second item in $n-1$ ways, whatever the choice of the first item, . . . , and the $r$ th item in $n-(r-1)$ ways, whatever the choice of the first $r-1$ items. By the multiplication principle the $r$ items can be chosen in $n \times(n-1) \times \cdots \times(n-r+1)$ ways.
For a nonnegative integer $n$, we define $n !$ (read $n$ factorial) by
$$n !=n \times(n-1) \times \cdots \times 2 \times 1$$
with the convention that $0 !=1$. We may then write
$$P(n, r)=\frac{n !}{(n-r) !}$$

## 数学代写|组合学代写Combinatorics代考|Combinations of Sets

Let $r$ be a nonnegative integer. By an $r$-combination of a set $S$ of $n$ elements, we understand an unordered selection of $r$ of the $n$ objects of $S$. In other words, an $r$-combination of $S$ is a subset of $S$ consisting of $r$ of the $n$ objects of $S$-that is, an $r$-element subset of $S$. If $S=$ ${a, b, c, d}$, then
$${a, b, c},{a, b, d},{a, c, d},{b, c, d}$$
are the four 3-combinations of $S$. We denote by $\left(\begin{array}{l}n \ r\end{array}\right)$ the number of $r$-combinations of an $n$-element set. ${ }^5$ Obviously,
$$\left(\begin{array}{l} n \ r \end{array}\right)=0 \quad \text { if } r>n .$$
Also,
$$\left(\begin{array}{l} 0 \ r \end{array}\right)=0 \quad \text { if } r>0 .$$
The following additional facts are readily seen to be true for each nonnegative integer $n$ :
$$\left(\begin{array}{l} n \ 0 \end{array}\right)=1,\left(\begin{array}{l} n \ 1 \end{array}\right)=n,\left(\begin{array}{l} n \ n \end{array}\right)=1 .$$
In particular, $\left(\begin{array}{l}0 \ 0\end{array}\right)=1$. The basic formula for combinations is given in the next theorem.

## 数学代写|组合学代写Combinatorics代考|Permutations of Sets

$a b c$，
$S$的6种2-排列是
$a b \quad a c \quad b a \quad b c \quad c a \quad c b$，
$S$的6种3排列是
$a b c \quad a c b \quad b a c$ bca $c a b \quad c b a$。
$S$没有4个排列，因为$S$的元素少于4个。

$$P(n, r)=n \times(n-1) \times \cdots \times(n-r+1) .$$

$$n !=n \times(n-1) \times \cdots \times 2 \times 1$$

$$P(n, r)=\frac{n !}{(n-r) !}$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。