# Practice Quiz 3 Recursive Definitions Relations Basic Counting Pigeonhole Principle Permutations & Combinations Discrete Probability.

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Practice Quiz 3 Recursive Definitions Relations Basic Counting Pigeonhole Principle Permutations & Combinations Discrete Probability

Problems The Fibonacci numbers f 0, f 1, f 2, … are defined as f 0 = 0, f 1 = 1, and f n = f n-1 + f n-2, for n > 1. Prove that the function f(n) = f 1 + f 3 + … + f 2n-1 is equal to f 2n whenever n is a positive integer.

Consider the relation R = {x, y | x + y > 10} on the set of positive integers. –Is R reflexive? Justify your answer. –Is R symmetric? Justify your answer. –Is R antisymmetric? Justify your answer. –Is R transitive? Justify your answer. Suppose that R1 and R2 are symmetric relations on a set A. Prove or disprove that R1 – R2 (set difference) is also symmetric. Problems

Given the relation on the set {0, 1, 2, 3} defined by the ordered pairs (0,0), (1,1), (1,3), (2,2), (2,3), (3,1), (3,3) –What is the 0-1 matrix representation of this relation? –Is the relation reflexive, symmetric and/or transitive? –Is the relation an equivalence relation?

How many students must be in a class to guarantee that at least 5 were born on the same day of the week? What is the coefficient of x 7 y 12 in the expansion of (x+y) 19 ? What is the probability that a randomly selected day of the year (365 days) is in May? Suppose that you pick two cards one at a time at random from an ordianary deck of 52 cards. Find p(both cards are diamonds). Problems

The Fibonacci numbers, f 0, f 1, f 2, … are defined as f 0 = 0, f 1 = 1, and f n = f n-1 + f n-2, for n > 1. Prove that the function  (n) = f 1 + f 3 + … + f 2n-1 is equal to f 2n whenever n is a positive integer. Basis Step If n = 1, then  (1) = f 2*1 - 1 = f 1 = 1 = f 2.

Inductive Step Assume  (k) = f 1 + f 3 + … + f 2k-1 = f 2k for k  n. We must show that this implies that  (n+1) = f 2(n+1)  (n+1) = f 1 + f 3 + … + f 2n-1 + f 2n+1  (n+1) =  (n) + f 2n+1  (n+1) = f 2n + f 2n+1 = f 2n+2 = f 2(n+1) The Fibonacci numbers, f 0, f 1, f 2, … are defined as f 0 = 0, f 1 = 1, and f n = f n-1 + f n-2, for n > 1. Prove that the function  (n) = f 1 + f 3 + … + f 2n-1 is equal to f 2n whenever n is a positive integer.

Given the relation on the set {0,1,2,3} defined by the ordered pairs: (0,0), (0,2), (1,1), (1,2), (2,0), (2,2), (3,3). Is the relation: – Reflexive? Yes –Symmetric? No –Transitive? No –An equivalence relation? No What is the 0-1 matrix?

Suppose that R1 and R2 are symmetric relations on a set A. Prove or disprove that R1 – R2 (set difference) is also symmetric. Proof: Assume that R1 and R2 are symmetric relations on a set A. We must show that if (a,b)  R1 - R2, then (b,a)  R1 – R2. If (a,b)  R1 – R2, then (a,b)  R1 and (a,b)  R2. Since R1 is symmetric, (b,a)  R1. Since R2 is symmetric, it follows that (b,a)  R2, for if (b,a) was in R2, then (a,b) would have been in R2. Hence, since (b,a) is in R1 and not in R2, (b,a)  R1 – R2. Therefore R1 – R2 is symmetric.

Consider the relation R = {x, y | x + y > 10} on the set of positive integers. Is R reflexive? Justify your answer. Is R symmetric? Justify your answer. Is R antisymmetric? Justify your answer. Is R transitive? Justify your answer. R is not reflexive, since 1 + 1 < 10, so (1,1) is not in R. R is symmetric, since x + y > 10 means that y + x > 10, so (x, y)  R implies (y, x)  R. R is not antisymmetric, since (2, 10) and (10, 2)  R, but 2  10. R is not transitive since (2, 9)  R and (9, 3)  R, but (2, 3)  R because 2 + 3 < 10.

How many students must be in a class to guarantee that at least 5 were born on the same day of the week? By the pigeonhole principle: 7*4+1 = 29

What is the coefficient of x 12 y 7 in the expansion of (x+y) 19 ? Solution: Use the binomial expansion formula So the 8th coefficient is C(19,7) = 19!/7!12! = 50,388.

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