Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n...

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Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2

Transcript of Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n...

Page 1: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

Exam 1 Review

5.1, 6.1-7.4, 8.1, 8.2

Page 2: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

5.1 Proof by Mathematical InductionClaim: 2 divides n2+n whenever n is a positive integer.Proof:• BASIS• INDUCTIVE STEP– Assume:– Show:

Page 3: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

Basic Counting Rules- Ch. 6

• SUM rule (for +)• PRODUCT rule (for *)• INCLUSION/EXCLUSION• COMPLEMENT rule– number= total – opposite– Ex: password with at least 2 vowels =

total – (password with 0 or 1 vowels)

Page 4: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

Counting cases- from sec 6.1, 6.3, 6.51. Order matters, repetition allowed

Multiplication RuleEx: Social Security numbers

2. Order matters, repetition NOT allowedPermutations: P(n,r)= n!/(n-r)!Ex: number of ways to pick 1st, 2nd, 3rd from 30

3. Order DOESN’T matter, repetition allowedsection 6.5 (stars and bars; objects and dividers)n categories, n-1 dividers, r objectsC(n-1+r, r) = C(n-1+r, n-1)

4. Order DOESN’T matter, repetition NOT allowedCombinations: C(n,r)= n!/ [(n-r)!*r!]Ex: number of ways to pick a committee of 3 from 30C(n-1+r, r) = C(n-1+r, n-1)

5. Permutations with DUPLICATE items n! / (n1 ! * n2 ! * n3 ! *…)

Ex: how many ways to rearrange the letters in MISSISSIPPIEx: how many ways to put 3 identical Finite books, 4 identical Bio books, and 5 identical Calc books on a

shelf

Page 5: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

Basic probability rulesKnow these:• P(E)=|E|/|S|• 0<= P(E) <= 1• P(E ‘ ) = 1 – P(E)

• Know how to set up Bayes Thm tree diagrams

These will be given:• Bayes Theorem• P(X=k)= nCk * p k q n-k

• Mean or Expected value μ = E(x)=

• Special case: Binomial μ = E(x)=np; Standard deviation σ = √(npq)

Page 6: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

Sample Ch. 6 and 7 problems

Sample problems:1. Passwords can be comprised of letters or digits.

(uses sum, multiplication, complement rules)How many of them are:a) 4-6 charactersb) 4-5 characters, with exactly 1 digitc) 4-5 characters, with exactly 2 digitsd) 4-5 characters, with at least 2 digits

Page 7: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

…Probability

2. Which "type" of counting problems are these? (case 1,2,3,4, or 5?)

a)An ice cream parlor has 28 different flavors, 8 different kinds of sauce, and 12 toppings.i)In how many different ways can a dish of three scoops of ice cream be made where each flavor can be used more than once and the order of the scoops does not matter?ii)How many different kinds of small sundaes are there if a small sundae contains one scoop of ice cream, a sauce, and a topping?

Page 8: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

…Probability

b) How many ways are there to choose a dozen donuts from 20 varieties:i) if there are no two donuts of the same variety?ii) if all donuts are of the same variety?

iii) if there are no restrictions?iv) if there are at least two varieties?v) if there must be at least six blueberry-filled donuts?vi) if there can be no more than six blueberry-filled donuts?

Page 9: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

…Probability

c) A professor writes 20 multiple choice questions, each with possible answer a,b,c, or d, for a test. If the number of questions with a,b,c, and d as their answer is 8,3,4, and 5, respectively, how many different answer keys are possible, if the questions can be placed in any order?

d) How many ways are there to assign 24 students to five faculty advisors?

e) A witness to a hit and run accident tells the police that the license plate of the car in the accident, which contains three letters followed by three digits, starts with the letters AS and contains both the digits 1 and 2. How many different license plates can fit this description?

Page 10: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

f) There are 7 types of bagels at the store. i) How many different ways could you pick 12 of them

and bring them to a meeting? ii) How many different ways could you choose to

select bagels to each on 12 consecutive days?g)How many ways could we rearrange 13 books on a

bookshelf:i)if all are different?ii)if 4 are identical chemistry books, 6 physics, and 3 math?

Page 11: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

h)How many ways could I there be to select 6 students out of 20 to receive A's?

i)How many ways could I guess who in this class will get the best, second, and third score on the exam?

j) How many ways can I select 3 women and 3 men from a Math Team (of 20 females and 25 males) to go to the National Math Tournament?

Page 12: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

3. Number of solutions: How many nonnegative solutions are there to x1

+ x2 + x3 = 30, where x1>1, x2>4, x3>2?

Page 13: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

4. how many bit strings of length 8: i) have at least 6 zero‘s?

ii) start with 10 and end with 010?

Page 14: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

Ch. 7: Probability

Basic Def• Know def of P(E) • Basic examples: dice, coins, cards,…– Ex: when tossing 2 dice, find P(sum is 11)=

• Know P(E’)= ______• Bayes will be given- be able to write out tree

Page 15: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

8.1- Types of Recur relation examples Examples: Given a recurrence relation, calculate a

term, say a6

Prove: an=n! is a solution to an=n*an-1, a0=1

Find a solution to an=n*an-1, a0=1

Applications: number of bit strings, number of ways to walk up the steps, Tower of Hanoi

Page 16: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

8.2– Thm. 1

Thm. 1: Let c1, c2 be elements of the real numbers. Suppose r2-c1r –c2=0 has two distinct roots r1 and r2,

Then the sequence {a n} is a solution of the recurrence relation an = ____________

iff an= __________ for n=0, 1, 2… where______

Page 17: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

Thm. 1 for two roots

Theorem 1: Let c1, c2 be elements of the real numbers.

Suppose r2-c1r –c2=0 has two distinct roots r1 and r2,

Then the sequence {a n} is a solution of the recurrence relation an = c1an-1 + c2 an-2

iff an=α1r1n+ α2r2

n for n=0, 1, 2… where α1 and α2

are constants.

Page 18: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

Steps for Solving 2nd degree LHRR-K

For degree 2: the characteristic equation is r2-c1r –c2=0 (roots are used to find explicit formula)• Find characteristic equation• Find roots• Basic Solution: an=α1r1

n+ α2r2n where r1 and r2

are roots of the characteristic equation• Solve for α1,α2 to find solution• Prove this is the solution

Page 19: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

Thm. 2 for one root

Theorem 2: Let c1, c2 be elements of the real numbers.

Suppose r2-c1r –c2=0 has only one root r0 ,

Then the sequence {a n} is a solution of the recurrence relation an = c1an-1 + c2 an-2 iff an=α1r0

n+ α2 n r0n

for n=0, 1, 2… where α1 and α2 are constants.

Page 20: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

Ex: 6. an =8an-1 -16an-2 for n≥2; a0=2 and a1=20.

Find characteristic equation Find solution

Page 21: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

Ex: 6. an =8an-1 -16an-2 for n≥2; a0=2 and a1=20.

• Prove the solution you just found is a solution

Page 22: Exam 1 Review 5.1, 6.1-7.4, 8.1, 8.2. 5.1 Proof by Mathematical Induction Claim: 2 divides n 2 +n whenever n is a positive integer. Proof: BASIS INDUCTIVE.

Formulas that will be GIVEN on Exams• Binomial• P(X=k)= nCk * p k q n-k

• Mean or Expected value μ = np

• Standard deviation σ = √(npq)

• Bayes’ Theorem• Summation =

• 8.2: outline for Thm. 1: Let c1, c2 be elements of the real numbers. Suppose r2-c1r –c2=0 has two distinct roots r1 and r2,

Then the sequence {a n} is a solution of the recurrence relation an = ____________

iff an= __________ for n=0, 1, 2… where______

(you fill in the gaps on Thm 1)

-----------------------------------• Know all other other relevant formulas