19th Ave New York, NY 95822, USA

# 数学代写代考| Permutations and Combinations 离散数学

## 数学代写| Permutations and Combinations 离散代考

my-assignmentexpert愿做同学们坚强的后盾，助同学们顺利完成学业，同学们如果在学业上遇到任何问题，请联系my-assignmentexpert™，我们随时为您服务！

## 离散数学代写

A permutation is an arrangement of a given number of objects, by taking some or all of them at a time. A combination is a selection of a number of objects where the order of the selection is unimportant. Permutations and combinations are defined in terms of the factorial function, which was defined in Chap. 4 .
Principles of Counting
(a) Suppose one operation has $m$ possible outcomes and a second operation has $n$ possible outcomes, then the total number of possible outcomes when performing the first operation followed by the second operation is $m \times n$ (Product Rule).
(b) Suppose one operation has $m$ possible outcomes and a second operation has $n$ possible outcomes, then the total number of possible outcomes of the first operation or the second operation is given by $m+n$ (Sum Rule).
Example (Counting Principle $(a)$ )
Suppose a dice is thrown and a coin is then tossed. How many different outcomes are there and what are they?
Solution
There are six possible outcomes from a throw of the dice, $1,2,3,4,5$ or 6 , and there are two possible outcomes from the toss of a coin, $\mathrm{H}$ or $\mathrm{T}$. Therefore, the total number of outcomes is determined from the product rule as $6 \times 2=12$. The outcomes are given by
$(1, \mathrm{H}),(2, \mathrm{H}),(3, \mathrm{H}),(4, \mathrm{H}),(5, \mathrm{H}),(6, \mathrm{H}),(1, \mathrm{~T}),(2, \mathrm{~T}),(3, \mathrm{~T}),(4, \mathrm{~T}),(5, \mathrm{~T}),(6, \mathrm{~T})$
Example (Counting Principle $(b))$
Suppose a dice is thrown and if the number is even a coin is tossed and if it is odd then there is a second throw of the dice. How many different outcomes are there?
Solution
There are two experimehts involved with the first experiment involving an even number and a toss of a coin. There are 3 possible outcomes that result in an even number and 2 outcomes from the toss of a coin. Therefore, there are $3 \times 2=6$ outcomes from the first experiment.

The second experiment involves an odd number from the throw of a dice and the further throw of the dice. There are 3 possible outcomes that result in an odd number and 6 outcomes from the throw of a dice. Therefore, there are $3 \times 6=18$ outcomes from the second experiment.
5.7 Permutations and Combinations
97
Finally, there are 6 outcomes from the first experiment and 18 outcomes from the second experiment, and so from the sum rule there are a total of $6+18=24$ outcomes.
Pigeonhole Principle
The pigeonhole principle states that if $n$ items are placed into $m$ containers (with $n>m$ ), then at least one container must contain more than one item (Fig. 5.1).
Examples (Pigeonhole Principle)
(a) Suppose there is a group of 367 people, then there must be at least two people with the same birthday.

This is clear as there are 365 days in a year (with 366 days in a leap year), and so as there are at most 366 possible birthdays in a year. The group size is 367 people, and so there must be at least two people with the same birthday.
(b) Suppose that a class of 102 students is assessed in an examination (the outcome from the exam is a mark between 0 and 100 ). Then, there are at least two students who receive the same mark.
This is clear as there are 101 possible outcomes from the test (as the mark that a student may achieve is between is between 0 and 100 ), and as there are 102 students in the class and 101 possible outcomes from the test, then there must be at least two students who receive the same mark.

$$P=\frac{A}{(1+r)^{n}}$$
5.6 货币和年金的时间价值
95

$$\frac{A}{(1+r)}+\frac{A}{(1+r)^{2}}+\cdots+\frac{A}{(1+r)^{n}}$$

$$\mathrm{PV}=\frac{A}{r}\left[1-\frac{1}{(1+r)^{n}}\right]$$

$$\开始{对齐} \mathrm{PV} &=\frac{1000}{0.1}\left[1-\frac{1}{(1.1)^{5}}\right] \ &=10000(0.3791) \ &=\ 3791 \end{对齐}$$

\begin{表格}{|l|l|l|l|}
\hline Table $5.2$ 年金现值&年份&金额&现值计算$(r=0.1)$\
\hline
\end{表格}

## 图论代考

(a) 假设一个操作有 $m$ 个可能的结果，而第二个操作有 $n$ 个可能的结果，那么执行第一个操作后执行第二个操作时可能结果的总数是 $m \times n$ (Product Rule )。
(b) 假设一个操作有 $m$ 个可能的结果，而第二个操作有 $n$ 个可能的结果，那么第一个操作或第二个操作的可能结果总数由 $m+n$ 给出（求和规则） .

$(1, \mathrm{H}),(2, \mathrm{H}),(3, \mathrm{H}),(4, \mathrm{H}),(5, \mathrm{H}) ,(6, \mathrm{H}),(1, \mathrm{~T}),(2, \mathrm{~T}),(3, \mathrm{~T}),(4, \mathrm{ ~T}),(5, \mathrm{~T}),(6, \mathrm{~T})$

5.7 排列组合
97

(a) 假设有一组 367 人，那么必须至少有两个人的生日相同。

(b) 假设有 102 名学生参加了一次考试（考试的结果是 0 到 100 之间的分数）。然后，至少有两名学生获得相同的分数。

## 密码学代考

• Cryptosystem
• A system that describes how to encrypt or decrypt messages
• Plaintext
• Message in its original form
• Ciphertext
• Message in its encrypted form
• Cryptographer
• Invents encryption algorithms
• Cryptanalyst
• Breaks encryption algorithms or implementations

## 编码理论代写

1. 数据压缩（或信源编码
2. 前向错误更正（或信道编码
3. 加密编码
4. 线路码