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这是密西西比大学 组合学课程的代写成功案例。
Math401课程简介
This course explores the counting methods of Math 301 in further depth. Additional topics may include Ramsey theory, combinatorial designs, graph theory, matroid theory, and other related topics.
3 Credits
PREREQUISITES
Math 301: Discrete Mathematics (Minimum grade: C)
Pre-Requisite: 24 Earned Hours
Prerequisites
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Math401 Combinatorics HELP（EXAM HELP， ONLINE TUTOR）
How many 6 card hands contain 3 pairs?
Choose values for pairs AND choose suits for the lowest value pair AND choose suits for the middle value pair AND choose suits for the highest value pair. This gives $\left(\begin{array}{c}13 \ 3\end{array}\right)\left(\begin{array}{c}4 \ 2\end{array}\right)^3=61,776$.
How many 5 card hands contain a straight? A straight is 5 consecutive cards from the sequence $\mathrm{A}, 2,3,4,5,6,7,8,9,10, \mathrm{~J}, \mathrm{Q}, \mathrm{K}, \mathrm{A}$ without regard to suit.
Choose the lowest value in the straight (A to 10) AND choose a suit for each of the 5 values in the straight. This gives $10 \times 4^5=10240$.
Although the previous answer is acceptable, a poker player may object since a “straight flush” is better than a straight-and we included straight flushes in our count. Since a straight flush is a straight all in the same suit, we only have 4 choices of suits for the cards instead of $4^5$. Thus, there are $10 \times 4=40$ straight flushes. Hence, the number of straights which are not straight flushes is $10240-40=10200$.
How many rearrangements of the letters in EXERCISES are there? How many arrangements of eight letters can be formed using the letters in EXERCISES? (No letter may be used more frequently than it appears in EXERCISES.)
1.3.7. This is like Exercise 1.2.3, but we’ll do it a bit differently. Note that EXERCISES contains 3 E’s, 2 S’s and 1 each of C, I, R and X. By the end of Example 1.18, we can use (1.4) with $N=9$, $m_1=3, m_2=2$ and $m_3=m_4=m_5=m_6=1$. This gives $9 ! / 3 ! 2 !=30240$.
It can also be done without the use of a multinomial coefficient as follows. Choose 3 of the 9 possible positions to use for the three E’s AND choose 2 of the 6 remaining positions to use for the two S’s AND put a permutation of the remaining 4 letters in the remaining 4 places. This gives us $\left(\begin{array}{c}9 \ 3\end{array}\right) \times\left(\begin{array}{c}6 \ 2\end{array}\right) \times 4$ !.
The number of eight letter arrangements is the same. To see this, consider a 9-list with the ninth position labeled “unused.”
Return to choosing teams (Example 1.21). Suppose half the people are women and half are men, that each team must be as nearly evenly split as possible, and that there is one referee of each sex. How many ways can this be done?
Think of the teams as labeled and suppose Teams 1 and 2 each contain 3 men. We can divide the men up in $\left(\begin{array}{c}11 \ 3,3,2,2,1\end{array}\right)$ ways and the women in $\left(\begin{array}{c}11 \ 2,2,3,3,1\end{array}\right)$ ways.
We must now count the number of ways to form the ordered situation from the unordered one. Be careful-it’s not $4 ! \times 2$ as it was in the example! Thinking as in the early card example, we start out two types of teams, say M or $\mathrm{F}$ depending on which sex predominates in the team. We also have two types of referees. Thus we have two $\mathrm{M}$ teams, two $\mathrm{F}$ teams, and one each of an $\mathrm{F}$ referee and an $\mathrm{M}$ referee. We can order the two $\mathrm{M}$ teams (2 ways) and the two $\mathrm{F}$ teams (2 ways), so there are only $2 \times 2$ ways to order and so the answer is $\left(\begin{array}{c}11 \ 3,3,2,2,1\end{array}\right)^2 \frac{1}{4}$.
MY-ASSIGNMENTEXPERT™可以为您提供CATALOG MATH401 COMBINATORICS组合学的代写代考和辅导服务！
