数学代写|MATH121A Combinatorics

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这是加州理工学院组合学课程的代写成功案例。

Math121a课程简介

Math 121 is our standard three-term course on combinatorics though we are only able to offer two terms this year. The first term covers “structural” combinatorics, the study of interesting discrete structures, both in terms of general properties as well as in terms of the existence or classification of particularly nice or extreme instances. We will spend much of the term on graphs and related topics such as Ramsey theory, with the remainder of the term spent covering other important structures Latin squares, codes, designs, finite geometries, etc.. The second term will cover more quantititative aspects of combinatorics, as well as applications to and from algebra.

Prerequisites 

Grades: Grades will be assessed on the basis of homeworks, which will be given roughly biweekly. There will be no exams (or midterms). Students may discuss problems with each other but have to write down the solutions independently.
The following restrictions hold for all problems of the set: Students may ask questions to the TA (if one exists) and the professor. Resources which you may use while working on the homework include any books and non-interactive websites (i.e. no posting of the questions on internet fora). For calculations which you can do by hand, you may use a computer algebra program Mathematica, Maple, etc.; you may also fell free to use a computer to gain intuition say by solving small instances of the problems, so long as your eventual proof does not depend on such a computation.  If you use significant ideas from any source, other than the books mentioned on this website, you should mention where you got them from.

Math121a Combinatorics HELP（EXAM HELP， ONLINE TUTOR）

1.1.1. We can form $n$ digit numbers by choosing the leftmost digit AND choosing the next digit AND $\cdots$ AND choosing the rightmost digit. The first choice can be made in 9 ways since a leading zero is not allowed. The remaining $n-1$ choices can each be made in 10 ways. By the Rule of Product we have $9 \times 10^{n-1}$.

To count numbers with at most $n$ digits, we could sum up $9 \times 10^{k-1}$ for $1 \leq k \leq n$. The sum can be evaluated since it is a geometric series. This does not include the number 0 . Whether we add 1 to include it depends on our interpretation of the problem’s requirement that there be no leading zeroes. There is an easier way. We can pad out a number with less than $n$ digits by adding leading zeroes. The original number can be recovered from any such $n$ digit number by stripping off the leading zeroes. Thus we see by the Rule of Product that there are $10^n$ numbers with at most $n$ digits. If we wish to rule out 0 (which pads out to a string of $n$ zeroes), we must subtract 1 .

List the elements of the set in any order: $a_1, a_2, \ldots, a_{|S|}$. We can construct a subset by including $a_1$ or not AND including $a_2$ or not AND including $a_{|S|}$ or not. Since there are 2 choices in each case, the Rule of Product gives $2 \times 2 \times \cdots \times 2=2^{|S|}$.

If we want all assignments of birthdays to people, then repeats are allowed in the list mentioned in the hint. This gives $365^{30}$. If we want all birthdays distinct, no repeats are allowed in the list. This gives $365 \times 364 \times \cdots \times(365-29)$. The ratio is 0.29 . How can this be computed? There are a lot of possibilities. Here are some.

• Use a symbolic math package.
• Write a computer program.
• Use a calculator. Overflow may be a problem, so you might write the ratio as $(365 / 365) \times(364 / 365) \times \cdots \times(336 / 365)$
• Use (1.2). You are asked to do this in the next problem. Unfortunately, there is no guarantee how large the error will be.
• Use Stirling’s formula after writing the numerator as 365 !/335!. Since Stirling’s formula has an error guarantee, we know we are close enough. Computing the values directly from Stirling’s formula may cause overflow. This can be avoided in various ways. One is to rearrange the various factors by using some algebra:
  $$
  \frac{\sqrt{2 \pi 365}(365 / e)^{365}}{\sqrt{2 \pi 335}(335 / e)^{335}(365)^{30}}=\sqrt{365 / 335}(365 / 335)^{335} / e^{30}
  $$
  Another way is to compute the logarithm of Stirling’s formula and use that to estimate the logarithm of the answer.

Each of the 7 letters ABMNRST appears once and each of the letters CIO appears twice. Thus we must form an ordered list from the 10 distinct letters. The solutions are
$$
\begin{array}{rlrl}
k & =2: & 10 \times 9 & =90 \
k & =3: & 10 \times 9 \times 8 & =720 \
k & =4: 10 \times 9 \times 8 \times 7 & =5040
\end{array}
$$

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