# 数学代写|组合学代写Combinatorics代考|MATH482

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

1. In the chapter about sets, we considered several special cases of this principle. Below, we present the general case and some typical applications of the inclusion-exclusion principle.
Knowing the amounts of elements $|A|$ and $|B|$ of (finite) sets $A$ and $B$ is not enough to define the number of elements of the union $A \cup B$ of these sets. The reason is on the surface: the sets $A$ and $B$ can share elements. Depending on the amount of common elements, the number $|A \cup B|$ may vary from $\max {|A|,|B|}$ (greater of the numbers $|A|$ and $|B|$ ) and $|A|+|B|$. For instance, if $A \subset B$, then $|A \cup B|=|B|$, as $A \cup B=B$. Alternatively, if $A \cap B=\emptyset$ ( $A$ and $B$ do not have common elements), then $|A \cup B|=|A|+|B|$. The more elements are there in the intersection $A \cap B$, the less elements compared to $|A|+|B|$ are there in the union $A \cup B$. In fact, the connection between the numbers $|A|,|B|,|A \cup B|$ and $|A \cap B|$ is expressed by the equality
$$|A \cup B|=|A|+|B|-|A \cap B| .$$
This formula is called the inclusion-exclusion principle for two (finite) sets. It is universal in the sense that it is always correct, disregarding the power of the intersection of $A$ and $B$. Fig. $5.1$ illustrates the inclusion-exclusion principle schematically. The shape hatched with ascending lines denotes the set $A$, and the one hatched with descending lines denotes the set $B$. The double-hatched shape denotes the set $A \cap B$. The number $|A|+|B|$ exceeds the power of the union $A \cup B$ by $|A \cap B|$, as adding the numbers of elements of the sets $A$ and $B$ we account their shared elements (that is, the elements of the intersection $A \cap B$ ) twice. Subtracting the number $|A \cap B|$ from $|A|+|B|$, we get the number which accounts for any element of the union once. This is what the inclusion-exclusion principle states.

The formula of inclusion-exclusion principle gets more complicated with the growth of the number of sets. For three sets $A, B$, and $C$, it looks as follows
$$|A \cup B \cup C|=(|A|+|B|+|C|)-(|A \cap B|+|B \cap C|+|C \cap A|)+(|A \cap B \cap C|) .$$
Now, the right-hand side contains the summands of three types. In order to stress this, we have surrounded the similar summands with parentheses. Between the first pair of parentheses, there is the sum of the number of elements of each set. The second pair of parentheses contains the sum of numbers of elements of pairwise intersections of the original sets. Finally, the number in the third pair of parentheses is the number of elements of the intersection of all three sets.

## 数学代写|组合学代写Combinatorics代考|Zigzags in a Circle without Self-Intersections

Let us take $n(n \geq 2)$ points on a circle. For our convenience, we further call them base points. Choose one of them to be the initial point and denote it by $A$. The rest of the base points are denoted with the letters $B, C, D$, and so on.

Example 6.1. Imagine that one needs to depart from the initial point $A$ and visit each of base points once, moving along chords of the circle. The journey ends in one of the base points. How many ways are there to perform this if the whole trajectory should consist of the chords connecting the base points, which do not intersect inside the circle?

For example, if there are 4 base points on a circle $(n=4)$, namely $A, B, C, D$ (see Fig.6.1), then there are four wanted paths (trajectories): $A B C D, A B D C, A C B D$ and $A C D B$.
Let us get down to the solution. Let $B$ and $C$ be the base points adjacent to the initial point $A$. This means that one of two arcs in which each of the points $B$ and $C$ split the circle into, there is no other base point except for $A$. A polygonal chain which is the subject of the question, can not begin with any line segment other than $A B$ or $A C$, as any other chord $A P$ splits the points $B$ and $C$, so after visiting one of them we inevitably cross the chord in order to visit the other. Hence, any polygonal chain should begin with one of the segments $A B$ or $A C$. These two cases are equivalent (or symmetrical) in the sense that it is impossible to imagine why there could be more polygonal chains beginning with the line segment $A B$ than those beginning with $A C$. If we denote the wanted amount by $l(n)$ (we remind that $n$ is the number of the base points), then
$$l(n)=2 \cdot l_{A B}(n),$$
where $l_{A B}(n)$ is the amount of polygonal chains beginning with the line segment $A B$. Subsequent line segments of every such polygonal chain form a polygonal chain with the starting point $B$, which includes all base points, except for $A$, and possess all the required properties. This means that
$$l_{A B}(n)=l(n-1),$$
hence,
$$l(n)=2 \cdot l(n-1) .$$

# 组合学代写

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

1. 在关于集合的章节中，我们考虑了这个原理的几个特殊情况。下面，我们将介绍包含排除原埋的一般 情况和一些典型应用。
了解元綁的数量 $|A|$ 和 $|B|$ (有限) 集合 $A$ 和 $B$ 不足以定义并集的元嗉个数 $A \cup B$ 这些夽。原因就在表 面上：集合 $A$ 和 $B$ 可以共享元素。根据共同元溸的数量，数量 $|A \cup B|$ 可能会有所不同 $\max |A|,|B|$ (较大的数字 $|A|$ 和 $|B|$ ) 和 $|A|+|B|$. 例如，如果 $A \subset B$ ，然后 $|A \cup B|=|B|$ ，作为 $A \cup B=B$ 或者，如果 $A \cap B=\emptyset$ ( $A$ 和 $B$ 没有共同的元青)，那么 $|A \cup B|=|A|+|B|$. 交叉点中的元溸越多 $A \cap B$ ，相比于更少的元素 $|A|+|B|$ 工会里有吗 $A \cup B$. 事实上，数字之间的联系 $|A|,|B|,|A \cup B|$ 和 $|A \cap B|$ 由等式表示
$$|A \cup B|=|A|+|B|-|A \cap B| .$$
这个公式被称为两个 (有限) 集合的包含 – 排除原则。它是普遍的，因为它总是正确的，忽略交叉的 力量 $A$ 和 $B$. 如图。5.1示意性地说明了包含 – 排除原则。带有上升线的阴影形状表示集合 $A$, 带有下降 线的阴影表示集合 $B$. 双影线形状表示集合 $A \cap B$. 号码 $|A|+|B|$ 超过工会的权力 $A \cup B$ 经过 $|A \cap B|$ ，作为添加集合的元嫊数 $A$ 和 $B$ 我们考虑它们的共享元素（即交集的元嫊 $A \cap B$ ) 两次。减去数字 $|A \cap B|$ 从 $|A|+|B|$ ，我们得到一次占并集任何元素的数字。这就是包含-排除原则所说的。
随着集合数的增加，包含排除原理的公式变得更加复杂。三组 $A, B$ ，和 $C$ ，看起来如下
$$|A \cup B \cup C|=(|A|+|B|+|C|)-(|A \cap B|+|B \cap C|+|C \cap A|)+(|A \cap B \cap C|) .$$
现在，右侧包含三种类型的和。为了强调这一点，我们用括号将相似的加法括起来。在第一对括号之 间，是每组元素个数的总和。第二对括号包含原始集合的成对交集的元絜数之和。最后，第三对括号中 的数字是所有三个集合的交集的元素个数。

## 数学代写|组合学代写Combinatorics代考|Zigzags in a Circle without Self-Intersections

$$l(n)=2 \cdot l_{A B}(n),$$

$$l_{A B}(n)=l(n-1),$$

$$l(n)=2 \cdot l(n-1) .$$

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