### 数学代写|概率论代写Probability theory代考|STAT4528

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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|概率论代写Probability theory代考|Combinations

suppose that we have ” $n$ ” distinct objects. The number of states of choosing ” $r$ ” distinct objects from these ” $n$ ” distinct objects (without considering the order of choices) is equal to:
$$C_r^n=\left(\begin{array}{l} n \ r \end{array}\right)=\frac{n !}{(n-r) ! \times(r !)}$$
To prove the above equation, it suffices to refer to one of the similar previous problems with a straightforward answer. The number of states of selecting ” $r$ ” distinct objects from ” $n$ ” distinct objects with consideration of their permutations (orders) is $P_r^n=\frac{n !}{(n-r) !}$, every $r !$ results of which are equivalent to one state of the new problem (selecting objects without consideration of the permutations). For instance, in choosing a three-member group from a ten-member group of people, every 3 ! states of the problem with consideration of the order of choices are equivalent to one state of the problem without consideration of the order of choices.

Therefore, the number of states that we can choose three of the seven distinct elements equals:
$$C_3^7=\left(\begin{array}{l} 7 \ 3 \end{array}\right)=\frac{P_3^7}{3 !}=\frac{7 !}{4 ! 3 !}$$
Likewise, in general, it can be shown that the number of states of choosing ” $r$ ” elements from the ” $n$ ” distinct elements is equal to:
$$C_r^n=\frac{P_r^n}{r !}=\frac{n !}{(n-r) ! r !}=\frac{(n)(n-1) \cdots(n-(r-1))}{r !}=\left(\begin{array}{l} n \ r \end{array}\right)$$

Suppose that a class consists of five boys and four girls.
a) How many ways can a group of size 3 be chosen from them?
b) How many ways can a group of size 3 consisting of one girl and two boys be chosen?
c) How many ways can a group of size 3 consisting of at most one boy be chosen?

## 数学代写|概率论代写Probability theory代考|Significant identities of the combinatorial topic

in this section, we are about to introduce some of the widely used combinatorial identities in the probability theory and prove them analytically. The first identity is as follows:
$$\left(\begin{array}{l} n \ r \end{array}\right)=\left(\begin{array}{c} n \ n-r \end{array}\right) ; \quad 0 \leq r \leq n$$
To prove it analytically, suppose we have an $n$-member set and we want to select ” $r$ ” members from them (left side of the identity). Such a selection can be made by firstly choosing $(n-r)$ members of the set, setting them aside (right side of the identity), and then regarding the remaining $r$ members as the leading members of the set.

The second combinatorial identity known as the Pascal’s identity is expressed as follows:
$$\left(\begin{array}{l} n \ r \end{array}\right)=\left(\begin{array}{c} n-1 \ r-1 \end{array}\right)+\left(\begin{array}{c} n-1 \ r \end{array}\right) ; \quad 1 \leq r \leq n$$
Consider an $n$-member set and suppose that we want to select ” $r$ ” members from the set (left side of the identity). To do so, regard a specific element such as “A” and divide all the possible states into two groups. The first group consists of the states in which the member ” $\mathrm{A}$ ” is among the ” $r$ ” members selected, and the second group consists of states in which the member ” $\mathrm{A}$ ” is not among the ” $r$ ” members selected (right side of the identity). The number of possible states in which the member “A” is selected equals $\left(\begin{array}{l}1 \ 1\end{array}\right)\left(\begin{array}{l}n-1 \ r-1\end{array}\right)$, and the number of possible states in which the member ” $\mathrm{A}$ ” is not selected equals $\left(\begin{array}{l}1 \ 0\end{array}\right)\left(\begin{array}{c}n-1 \ r\end{array}\right)$. Hence, the total number of states is equal to:
$$\left(\begin{array}{l} 1 \ 0 \end{array}\right)\left(\begin{array}{c} n-1 \ r \end{array}\right)+\left(\begin{array}{l} 1 \ 1 \end{array}\right)\left(\begin{array}{l} n-1 \ r-1 \end{array}\right)=\left(\begin{array}{c} n-1 \ r \end{array}\right)+\left(\begin{array}{l} n-1 \ r-1 \end{array}\right)$$

# 概率论代考

## 数学代写|概率论代写Probability theory代考|Combinations

$$C_r^n=(n r)=\frac{n !}{(n-r) ! \times(r !)}$$

$$C_3^7=(73)=\frac{P_3^7}{3 !}=\frac{7 !}{4 ! 3 !}$$

$$C_r^n=\frac{P_r^n}{r !}=\frac{n !}{(n-r) ! r !}=\frac{(n)(n-1) \cdots(n-(r-1))}{r !}=(n r)$$

a) 有多少种方法可以从中选出一组大小为 3 的方法?
b) 一组由一个女孩和两个男孩组成的大小为 3 的小组有多少种选择?
c) 最多由一个男孩组成的 3 人组有多少种选择?

## 数学代写|概率论代写Probability theory代考|Significant identities of the combinatorial topic

$$(n r)=(n n-r) ; \quad 0 \leq r \leq n$$

$$(n r)=(n-1 r-1)+(n-1 r) ; \quad 1 \leq r \leq n$$

$$(10)(n-1 r)+(11)(n-1 r-1)=(n-1 r)+(n-1 r-1)$$

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## MATLAB代写

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