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数学代写|离散数学代写Discrete Mathematics代考|MATH2450 The Approximate Number of Subsets

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数学代写|离散数学代写Discrete Mathematics代考|The Approximate Number of Subsets

So, we know that the number of subsets of a 100 -element set is $2^{100}$. This is a large number, but how large? It would be good to know, at least, how many digits it will have in the usual decimal form. Using computers, it would not be too hard to find the decimal form of this number $\left(2^{100}=\right.$ 1267650600228229401496703205376 ), but suppose we have no computers at hand. Can we at least estimate the order of magnitude of it?

We know that $2^3=8<10$, and hence (raising both sides of this inequality to the 33rd power) $2^{99}<10^{33}$. Therefore, $2^{100}<2 \cdot 10^{33}$. Now $2 \cdot 10^{33}$ is a 2 followed by 33 zeros; it has 34 digits, and therefore $2^{100}$ has at most 34 digits.

We also know that $2^{10}=1024>1000=10^3$; these two numbers are quite close to each other ${ }^2$. Hence $2^{100}>10^{30}$, which means that $2^{100}$ has at least 31 digits.

This gives us a reasonably good idea of the size of $2^{100}$. With a little more high-school math, we can get the number of digits exactly. What does it mean that a number has exactly $k$ digits? It means that it is between $10^{k-1}$ and $10^k$ (the lower bound is allowed, the upper is not). We want to find the value of $k$ for which
$$10^{k-1} \leq 2^{100}<10^k .$$

数学代写|离散数学代写Discrete Mathematics代考|Sequences

Motivated by the “encoding” of subsets as strings of 0’s and 1’s, we may want to determine the number of strings of length $n$ composed of some other set of symbols, for example, $a, b$ and $c$. The argument we gave for the case of 0’s and 1’s can be carried over to this case without any essential change. We can observe that for the first element of the string, we can choose any of $a, b$ and $c$, that is, we have 3 choices. No matter what we choose, there are 3 choices for the second element of the string, so the number of ways to choose the first two elements is $3^2=9$. Proceeding in a similar manner, we get that the number of ways to choose the whole string is $3^n$.

In fact, the number 3 has no special role here; the same argument proves the following theorem:

Theorem 1.5.1 The number of strings of length $n$ composed of $k$ given elements is $k^n$.

The following problem leads to a generalization of this question. Suppose that a database has 4 fields: the first, containing an 8-character abbreviation of an employee’s name; the second, M or F for sex; the third, the birthday of the employee, in the format mm-dd-yy (disregarding the problem of not being able to distinguish employees born in 1880 from employees born in 1980 ); and the fourth, a job code that can be one of 13 possibilities. How many different records are possible?

The number will certainly be large. We already know from theorem 1.5.1 that the first field may contain $26^8>200,000,000,000$ names (most of these will be very difficult to pronounce, and are not likely to occur, but let’s count all of them as possibilities). The second field has 2 possible entries. The third field can be thought of as three separate fields, having 12,31 , and 100 possible entries, respectively (some combinations of these will never occur, for example, 04-31-76 or 02-29-13, but let’s ignore this). The last field has 13 possible entries.

Now how do we determine the number of ways these can be combined? The argument we described above can be repeated, just “3 choices” has to be replaced, in order, by ” $26^8$ choices,” “2 choices,” “12 choices,” “31 choices,” “100 choices,” and “13 choices.” We get that the answer is $26^8$. $2 \cdot 12 \cdot 31 \cdot 100 \cdot 13=201,977,536,857,907,200$.

We can formulate the following generalization of Theorem 1.5.1 (the proof consists of repeating the argument above).

$$数学代写|离散数学代写Discrete Mathematics代考|Sequences 10^{k-1} \leq 2^{100}<10^k .$$ 受将子集”编码”为 0 和 1 的字符串的启发，我们可能想要确定长度为 $n$ 由一些其他符号集组成，例如， $a, b$ 和 $c$. 我们针对 0 和 1 的 情况给出的论证可以在没有任何本质变化的情况下转移到这种情况。我们可以观暌到，对于字符串的第一个元耒，我们可以选择任 何一个 $a, b$ 和 $c$ ，也就是说，我们有 3 个选择。无诋我们选择什么，字符串的第二个元粖都有 3 种选择，所以选择前两个元溸的方式 数是 $3^2=9$. 以类似的方式进行，我们得到选择整个字符串的方法数是 $3^n$. 事实上，数字 3 在这里并没有什么特别的作用; 同样的论证证明了下面的定理: 定理1.5.1 length的串数 $n$ 由…组成的 $k$ 给定的元雑是 $k^n$. 以下问题导致了这个问题的概括。假设一个数据库有 4 个字段：第一个，包含员工姓名的 8 个字符的怕宿写；第二个，M或代表性 别; 第三，员工生日，格式为 $m m-d d-y y$ (忽略无法区分 1880 年和1980年出生员工的问题)；第四，工作代码可以是 13 种可能 性之一。有多少种不同的记录是可能的? 数量肯定会很大。我们已泾从定理 $1.5 .1$ 知道第一个字段可能包含 $26^8>200,000,000,000$ 名字（其中大部分都很难发音，而且 不太可能出现，但让我们把它们都算作可能性)。第二个字段有 2 个可能的条目。第三个字段可以被认为是三个单独的字段，分 别有 12,31 和 100 个可能的条目（这些的某些组合永远不会出现，例如 04-31-76 或 02-29-13，但让我们分略这个）。最后一个 字段有 13 个可能的条目。

$2 \cdot 12 \cdot 31 \cdot 100 \cdot 13=201,977,536,857,907,200$.

MATLAB代写

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