### 数学代写|组合学代写Combinatorics代考|CS519

statistics-lab™ 为您的留学生涯保驾护航 在代写组合学Combinatorics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写组合学Combinatorics代写方面经验极为丰富，各种代写组合学Combinatorics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|组合学代写Combinatorics代考|Bijection. Combinatorial Bijection Principle

Suppose that 59 teams are participating in a soccer cup. How many matches will be played? Even after additional explanations regarding the rules of the tournament, a large number of respondents hesitated to provide the answer, attempting the construction of various schemes and the related calculations. There were mathematicians among those who got confused about this issue, not to mention those who participate in the competition schedule. This is a kind of question to which the student can give an instant and reasonable answer, and at the same time it can make the specialist lose his balance and dig deep in search of the truth that is right on the surface. A foreword regarding the rules of cup competitions is needed. The classic system is that each match should end effectively (that is, by the victory of one of the teams), and the team that loses is no longer taking part in the tournament. This is the fundamental rule of the winner’s detection system, which is called a single-elimination, knockout, or sudden death tournament. The rest of the rules are not significant. Therefore, they are the responsibility of organizers of the competition (for example, football association). The organizers compile a schedule of the tournament, providing the rules for the creation of pairs at different stages of the competition, decide on which stage one or another team enter the tournament etc. They can also make a decision that the teams should play two matches on each stage instead of one. This does not change the essence of the knockout system, provided that after these two matches one of the two teams necessarily leaves the tournament. This alternative rule does not change our task either: the answer is simply doubled.

Therefore, assume we have a “classic competition”, when two teams play one match to determine which one of them is eliminated. How many matches will have to be played by all the teams?

The one, who focuses from the beginning on the various options of the schedule of competition, will waste a good deal of time searching for the answer. And this is the most popular route to a solution. Alternatively, the one, who realizes that the schedule of the competition is irrelevant to the task, no matter how simple or tricky it is, will get the answer almost immediately. The only important rule is the following: the losing team is eliminated from the competition. Imagine that the tournament is over. Which teams have not been knocked out? Only the cup winner. All the rest were eventually defeated and left the competition. There are 58 of them. And there were the same amount of matches, because each team lost in a single match, and each match resulted in a defeat of one of the teams. The teams, which lost in the tournament, are in such connection with the matches played, that there is no doubt that the number of matches and the number of losing teams are the same. This connection is called bijective correspondence (or one-to-one correspondence). We will have to deal with many more similar situations and use the term “bijective correspondence” or simply “bijection”, and therefore, it is time to stop and explain in detail its exact meaning.

## 数学代写|组合学代写Combinatorics代考|Bijection between paths

1. Here we will deal with the summation of numbers and vectors. If one needs to calculate the sum of several (many) summands, then by the appropriate positioning of parentheses, this task can be reduced to the repeated summation of two summands. Moreover, the parentheses can be positioned in many different ways. The result does not depend on this. This is one of the fundamental arithmetical laws. It can be deduced from the associativity of addition, which refers to any three summands. The reader is well familiar with this property from the elementary school. Symbolically, it is presented as follows:
$$(a+b)+c=a+(b+c) .$$
Considering the sums of many summands we will adhere to the following rule: each “+” sign must correspond to a certain pair of parentheses (opening and closing parentheses). Hence, there should be the same amount of pairs of parentheses as the amount of “+” signs in the expression. In particular, under such agreement, the associativity property is expressed as follows:
$$((a+b)+c)=(a+(b+c)) .$$
Actually, we are not interested in associativity law and its consequences. We are dealing with a purely combinatorial problem: how many ways are there to place parentheses correctly in the sum of $n$ summands? The word “correctly” here means that there should be
2. equal amounts of opening and closing parentheses, and every pair of parentheses (opening parenthesis; closing parenthesis) corresponds to a certain “+” sign. In other words, pairs of parentheses (opening and closing) must be in bijective correspondence with the “+” signs.
3. In the case of three summands, there are two ways to place parentheses: $((a+b)+c)$ and $(a+(b+c))$.

## 数学代写|组合学代写Combinatorics代考|Bijection between paths

1. 这里我们将处理数字和向量的求和。如果需要计算几个 (许多) 被加数的和，那么通过 括号的适当定位，这个任务可以简化为两个被擞的重复求和。此外，括号可以以许多 不同的方式放置。结果不取决于此。这是基本的算术定律之一。它可以从加法的结合性 推导出来，它指的是任何三个被吅数。读者从小学就熟悉这个属性了。象征性地，它呈 现如下:
$$(a+b)+c=a+(b+c) .$$
考虑到许多加数的总和，我们将遵循以下规则: 每个” ${ }^{\prime \prime}$ 号必须对应于特定的一对括号 (左括号和右括号) 。因此，括号对的数量应该与表达式中““”号的数量相同。特别地， 在这种约定下，结合性表示如下:
$$((a+b)+c)=(a+(b+c)) .$$
实际上，我们对结合律及其后果不感兴趣。我们正在处理一个纯粹的组合问题: 有多少 种方法可以正确地在总和中放置括号 $n$ 求和? 这里的“正确”一词意味着应该有
2. 等量的左右括号，每对括号 (左括号; 右括号) 对应某个“+”号。换句话说，括号对（左 括号和右括号）必须与“”“符号双射对应。
3. 在三个被加数的情况下，有两种放置括号的方法: $((a+b)+c)$ 和 $(a+(b+c))$.

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。