[SOLVED] CS代考计算机代写 Chapter 1 – Counting / Combinatorics

Chapter 1 – Counting / Combinatorics
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EX A
EX B EX C EX D
EX E EX F
EX G EX H
EX I EX J
EX K
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Ice cream. There are 5 flavors, served in two types of cone, with any combination of 3 toppings (choc dip; sprinkles; nuts). Let’s consider one scoop cones. (Should we include servings in a dish / no cone?) How many orders are possible?
More ice cream. How about two scoop cones? (Forget about the no-cone / in-a-dish option on this one.) Can the scoops be of different flavors? If so, does the order of the scoops matter?
6 people walk into a bike shop. 9 different bikes are for rent. In how many ways can each of the 6 people select bikes to ride.
8 different commercials will air during a 1-hour television program. 4 of the commercials are for a brand of beer; the other 4 are for an insurance company.
a. In how many orders can the commercials be aired?
b. Suppose the two types (beer; insurance) must be alternated. How many ways?
c. Suppose the four commercials of each product must be shown consecutively. How many ways?
d. Suppose the four beer commercials must be shown consecutively. How many ways?
After hosting a charity event, 4 of the 20 people in an organization must be chosen to do cleaning chores. How many selections are possible?
A coin is tossed 14 times. How many sequences of results have exactly 9 Hs? (For example, one such sequence is HHHHHHHHHTTTTT. (And – this sequence is no more or less likely than any other sequence of 9 Hs and 5 Ts.)
20 people in an organization host 5 charity events. After each event, 4 people are selected to clean up. No one will clean up twice. How many selections of 5 groups of 4 to staff each of the 5 events are possible?
Consider a class of 34 students.
a. If the class is split in half: How many splits are possible?
b. Suppose the class is split into six groups to solve problems 1 through 6. Four groups of 6 are chosen to solve problems 1 – 4; 2 two groups of 5 solve 6 and 6. In how many ways can this be done?
Ice cream again. Three scoops – but (only) in a dish. There are still 3 toppings available.
Consider rearranging the letters in the word STATISTICS.
a. How many distinguishable 10-letter sequences (“words” of a sort) are possible?
b. How many are there such that the three S’s are not in sequence – no SSS? (Count first how many have the three S’s in sequence.)
c. How many of them have an S on both ends? (At least one does!)
12 juniors and 13 seniors are in an organization. 3 juniors and 4 seniors are to be selected to drive in a school van to attend a conference. How many such travel groups are possible if…
a. …there are two juniors who will not travel together
b. …there is a junior and a senior who will not travel together

EX L
In the card game cribbage, players are dealt 6 cards. Two of the cards are put aside for later use. The remaining 4 cards are played in order (alternating between players). Both parts – which cards to put aside and the order in which the remainder are played – have an impact on scoring. If, having been dealt the 6 cards, a player wishes to assess (or have a computer assess) all possible ways of playing the hand (to determine an optimal strategy), how many ways are considered? In other words: How many ways can 4 cards be sequenced.
From the text
[note: it will help to first determine how many selections are possible ignoring any enmity among students]
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EX A Observe a future annual high-water level of Lake Ontario.
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EX B Toss 4 coins (or a single coin 4 times).
EX C Deck of 6 cards labelled 1, 2, … , 6. Choose two cards (don’t worry about order) without replacement. (Note: If with replacement, order will necessarily become an issue.) We’ll let ij indicate the outcome that cards i and j are selected. (This is horrible – but minimal – notation.)
Sample Space
S = { ij : i = 1, 2, 3, 4, 5; j = (i + 1), … , 6 }
= { 12, 13, 14, 15, 16, 23, 24, 25, 26, 34, 35, 36, 45, 46, 56 }
Events:
A = { 12, 13, 14, 23, 24, 34 } B = { 13, 24, 35, 46 }
C = { 15, 24 } D = { 15, 25, 35, 45 } E = { 14, 24, 34 }
Axioms of Probability
Probability is a function that assigns a number to an event. Only three rules are needed to get things off the ground in a fashion that agrees with our intuitive notion of what probability should be.
1. For any event A, 0 ≤ P(A) ≤ 1.
2. P(S) = 1.
∞∞
3. For events A , A , that are mutually exclusive: P A = P(A ).
12 
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Implication (way down the road): Imagine the experiment is repeated many times where the starting conditions are identical each time and the results of any previous experiments have no impact on the way future experiments unfold. The relative frequency of occurrence of an event will with virtual certainty approach the probability of that event. A probability can be thought of as an ideal relative frequency.
i∑  i=1  i=1

Propositions
(2 and 4 do not appear in the text.) For events A and B.
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1) 2)
3)
4) 5)
P(A)+P(Ac)=1
P(A)= P(AB) P AB+ (This is event decomposition – decomposing A into the mutually exclusive
()
c
parts occurring with and not with B.) If A⊂Bthen P(A)≤P(B)
P(A∪B)= P(AB) P(+ABc ) P(+AcB). (More decomposition.) P(A∪B)=P(A) P+(B) P−(AB)
Note that Axiom 3 handles / simplifies 5 above for mutually exclusive events. It is simple to find probabilities of unions (“at least one”) for mutually exclusive events having known probabilities. It’s also often easy to compute probabilities for intersections such as AB (coming attraction). Many problems involve unions of mutually exclusive intersections such as in the items 2 and 4 above.
Venn Diagrams are useful for two and three events (maybe more?).
Inclusion / Exclusion
Two events: P(A∪B)= P(A) P+(B) P−(AB)
Three events: PABC PAPBPC PABPACPBC PABC
Four events:
PABCD PAPBPCPD 
In general for a finite collection of events E1, E2, … , En.
More succinctly.
 
 PAB PAC PAD PBC PBD PCD 
  PABC PABD PACD PBCD
 PABCD

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