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# 数学代写|随机过程Stochastic Porcess代考|PERMUTATIONS

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## 数学代写|随机过程代写Stochastic Porcess代考|BASIC THEOREMS

To count the number of outcomes of an experiment or the number of possible ways an event can occur, it is often useful to look for special patterns. Sometimes patterns help us develop techniques for counting. Two simple cases in which patterns enable us to count easily are permutations and combinations. We study these two patterns in this section and the next.
Definition 2.1 An ordered arrangement of $r$ objects from a set A containing $n$ objects $(0<r \leq n)$ is called an r-element permutation of $A$, or a permutation of the elements of A taken $r$ at a time. The number of $r$-element permutations of a set containing $n$ objects is denoted by ${ }_n P_r$.

By this definition, if three people, Brown, Smith, and Jones, are to be scheduled for job interviews, any possible order for the interviews is a three-element permutation of the set {Brown, Smith, Jones}.

If, for example, $A={a, b, c, d}$, then $a b$ is a two-element permutation of $A, a c d$ is a three-element permutation of $A$, and $a d c b$ is a four-element permutation of $A$. The order in which objects are arranged is important. For example, $a b$ and $b a$ are considered different twoelement permutations, $a b c$ and $c b a$ are distinct three-element permutations, and $a b c d$ and $c b a d$ are different four-element permutations.

To compute ${ }_n P_r$, the number of permutations of a set $A$ containing $n$ elements taken $r$ at a time $(1 \leq r \leq n)$, we use the generalized counting principle: Since $A$ has $n$ elements, the number of choices for the first object in the $r$-element permutation is $n$. For the second object, the number of choices is the remaining $n-1$ elements of $A$. For the third one, the number of choices is the remaining $n-2, \ldots$, and, finally, for the $r$ th object the number of choices is $n-(r-1)=n-r+1$. Hence
$${ }_n P_r=n(n-1)(n-2) \cdots(n-r+1) .$$
An $n$-element permutation of a set with $n$ objects is simply called a permutation. The number of permutations of a set containing $n$ elements, ${ }_n P_n$, is evaluated from (2.1) by putting $r=n$.
$${ }_n P_n=n(n-1)(n-2) \cdots(n-n+1)=n ! .$$

## 数学代写|随机过程代写Stochastic Porcess代考|COMBINATIONS

In many combinatorial problems, unlike permutations, the order in which elements are arranged is immaterial. For example, suppose that in a contest there are 10 semifinalists and we want to count the number of possible ways that three contestants enter the finals. If we argue that there are $10 \times 9 \times 8$ such possibilities, we are wrong since the contestants cannot be ordered. If $A$, $B$, and $C$ are three of the semifinalists, then $A B C, B C A, A C B, B A C, C A B$, and $C B A$ are all the same event and have the same meaning: ” $A, B$, and $C$ are the finalists.” The technique known as combinations is used to deal with such problems.

Definition 2.2 An unordered arrangement of $r$ objects from a set A containing $n$ objects $(r \leq n)$ is called an $r$-element combination of $A$, or a combination of the elements of A taken $r$ at a time.

Therefore, two combinations are different only if they differ in composition. Let $x$ be the number of $r$-element combinations of a set $A$ of $n$ objects. If all the permutations of each $r$-element combination are found, then all the $r$-element permutations of $A$ are found. Since for each $r$-element combination of $A$ there are $r$ ! permutations and the total number of $r$-element permutations is ${ }_n P_r$, we have
$$x \cdot r !={ }_n P_r .$$
Hence $x \cdot r !=n ! /(n-r) !$, so $x=n ! /[(n-r) ! r !]$. Therefore, we have shown that
The number of $r$-element combinations of $n$ objects is given by
$${ }_n C_r=\frac{n !}{(n-r) ! r !} \text {. }$$
Historically, a formula equivalent to $n ! /[(n-r) ! r !]$ turned up in the works of the Indian mathematician Bhaskara II (1114-1185) in the middle of the twelfth century. Bhaskara II used his formula to calculate the number of possible medicinal preparations using six ingredients. Therefore, the rule for calculation of the number of $r$-element combinations of $n$ objects has been known for a long time.

It is worthwhile to observe that ${ }_n C_r$ is the number of subsets of size $r$ that can be constructed from a set of size $n$. By Theorem 2.3, a set with $n$ elements has $2^n$ subsets. Therefore, of these $2^n$ subsets, the number of those that have exactly $r$ elements is ${ }_n C_r$.

## 数学代写|随机过程代写Stochastic Porcess代考|BASIC THEOREMS

$${ }_n P_r=n(n-1)(n-2) \cdots(n-r+1) .$$
a $n$的集合的-元素置换 $n$ 对象被简单地称为排列。集合中包含的排列的数目 $n$ 元素， ${ }_n P_n$，由式(2.1)求值 $r=n$．
$${ }_n P_n=n(n-1)(n-2) \cdots(n-n+1)=n ! .$$

## 数学代写|随机过程代写Stochastic Porcess代考|COMBINATIONS

$$x \cdot r !={ }_n P_r .$$

$n$对象的$r$元素组合的个数由
$${ }_n C_r=\frac{n !}{(n-r) ! r !} \text {. }$$

## MATLAB代写

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