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Exam 1 Review

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Basic Counting Rules- ch. 5 SUM rule (for +) PRODUCT rule (for *) INCLUSION/EXCLUSION COMPLEMENT rule – number= total – opposite – Ex: number with at least 2 vowels = total – (number with 0 or 1 vowels)

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5.1, 5.3, 5.5 Order matters, repetition allowed – Multiplication Rule – Ex: Social Security numbers Order matters, repetition NOT allowed – Permutations: P(n,r)= n!/(n-r)! – Ex: number of ways to pick 1 st, 2 nd, 3 rd from 30 Order DOESN’T matter, repetition allowed – section 5.5 (stars and bars; objects and dividers) – n categories, n-1 dividers, r objects – C(n-1+r, r) = C(n-1+r, n-1) Order DOESN’T matter, repetition NOT allowed – Combinations: C(n,r)= n!/ [(n-r)!*r!] – Ex: number of ways to pick a committee of 3 from 30

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Binomial P(X=k)= nCk * p k q n-k = np σ = (npq)

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Basic probability rules P(E)=|E|/|S| 0<= P(E) <= 1 P(E ‘ ) = 1 – P(E) For Bayes Thm: do tree diagram Expected value: E(X)= = for binomial = np

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Sample Ch. 5 and 6 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 characters b) 4-5 characters, with exactly 1 digit c) 4-5 characters, with exactly 2 digits d) 4-5 characters, with at least 2 digits

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…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?

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…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?

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…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?

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… 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?

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… 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 female mathletes and 25 male mathletes) to go to the National Math Tournament?

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… 3. Number of solutions a) How many nonnegative solutions are there to x 1 + x 2 + x 3 = 30, where x 1 >1, x 2 >4, x 3 >2?

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… 7. how many bit strings of length 8: i) have at least 6 zero‘s? ii) start with 10 and end with 010?

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… 8. a) How many one-to-one functions exist from a set with 3 elements to a set with 7 elements (section 5.1)?

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Ch. 6: Probability Basic Def P(E’) Bayes

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7.1- Recurrence relations example Prove: a n =n! is a solution to a n =n*a n-1, a 0 =1 Find a solution to a n =n*a n-1, a 0 =1

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7.2– Thm. 1 Thm. 1: Let c 1, c 2 be elements of the real numbers. Suppose r 2 -c 1 r –c 2 =0 has two distinct roots r 1 and r 2, Then the sequence {a n } is a solution of the recurrence relation a n = ____________ iff a n = __________ for n=0, 1, 2… where______

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Thm. 2 Thm. 1: Let c 1, c 2 be elements of the real numbers. Suppose r 2 -c 1 r –c 2 =0 has ____root, Then the sequence {a n } is a solution of the recurrence relation a n = ____________ iff a n = __________ for n=0, 1, 2… where______

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Summation formula Given:=

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Solving 2 nd degree LHRR-K For degree 2: the characteristic equation is r 2 - c 1 r –c 2 =0 (roots are used to find explicit formula) Basic Solution: a n =α 1 r 1 n + α 2 r 2 n where r 1 and r 2 are roots of the characteristic equation

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Thm. 1 for two roots Theorem 1: Let c 1, c 2 be elements of the real numbers. Suppose r 2 -c 1 r –c 2 =0 has two distinct roots r 1 and r 2, Then the sequence {a n } is a solution of the recurrence relation a n = c 1 a n-1 + c 2 a n-2 iff a n =α 1 r 1 n + α 2 r 2 n for n=0, 1, 2… where α 1 and α 2 are constants.

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Thm. 2 for one root Theorem 2: Let c 1, c 2 be elements of the real numbers. Suppose r 2 -c 1 r –c 2 =0 has only one root r 0, Then the sequence {a n } is a solution of the recurrence relation a n = c 1 a n-1 + c 2 a n-2 iff a n =α 1 r 0 n + α 2 n r 0 n for n=0, 1, 2… where α 1 and α 2 are constants.

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Ex: 6. a n =8a n-1 -16a n-2 for n≥2; a 0 =2 and a 1 =20. Find characteristic equation Find solution

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Ex: 6. a n =8a n-1 -16a n-2 for n≥2; a 0 =2 and a 1 =20. Prove the solution you just found is a solution

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