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

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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

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