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MATH579课程简介
Learning Objectives:
There are three distinct phases to solving a combinatorial problem. Generally, the first phase is the most difficult to learn, and the last phase is the easiest. Students will learn all three in this course. First, the problem must be categorized as to which combinatorial tool would be appropriate. Second, a model must be created that translates the abstract formulation of the problem into the symbols required for the combinatorial tools to work. ‘hird, the combinatorial tools must be applied to the symbols.
Prerequisites
Portfolio:
Students are expected to keep a portfolio in a three-ring binder or something similar, containing a detailed and complete solution to every exercise in the text (those marked + or ++ are optional). These portfolios will not be collected or checked, except upon a student’s request; however, they will be an invaluable resource in preparing for exams.
Students are NO’T’ required to personally solve every exercise appearing in their portfolios; they are strongly encouraged to collaborate with classmates. However, before accepting a classmate’s solution into their portfolio, students are expected to carefully check it for completeness and correctness.
MATH579 Combinatorics HELP（EXAM HELP， ONLINE TUTOR）
How many words are there of length $n$, with the first three letters vowels, and the remaining letters consonants?
There are $5^3$ three-letter words consisting entirely of vowels. There are $21^{n-3}$ words of length $n-3$, consisting entirely of consonants. We select one of each; hence the answer is $5^3 21^{n-3}$. Note that this is only valid for $n \geq 3$. For $n=1,2$, the answer is 0 .
How many words are there of length $n$, with exactly three $a$ ‘s?
If $n=1,2$, the answer is 0 . Now assume $n \geq 3$. We mark $n$ places where we will place letters. Three of them will be $a$ ‘s; there are $\left(\begin{array}{l}n \ 3\end{array}\right)$ ways to pick those three locations. Think of a set of size 3 drawn from $[n]$, where the set marks the locations where a’s will go.
The remaining $n-3$ spaces are filled consecutively with any word, drawn from the alphabet that contains no $a$ (equivalent to [25]). There are $25^{n-3}$ ways to do this. Combining, our solution is $\left(\begin{array}{l}n \ 3\end{array}\right) 25^{n-3}$.
How many words are there of length $n$, with no two consecutive letters being the same?
If $n<26$ then the answer is 0 . Otherwise the answer is $26 ! S(n, 26)$.
If $n=1,2$ the answer is 0 ; we assume $n \geq 3$. We again mark $n$ places for letters, and place our three $a$ ‘s. There are $n-2$ choices for the first $a$ (it can’t be in the last two spaces because then there wouldn’t be enough room for the other $a$ ‘s afterward); the remaining a’s will follow right afterwards. Then, we fill the remaining apces with non- $a$ letters as previously. Hence, the solution is $(n-2) 25^{n-3}$.
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