Practice Quiz 3

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

2. Problems • The Fibonacci numbers f0, f1, f2, … are defined as f0 = 0, f1 = 1, and fn = fn-1 + fn-2 , for n > 1. Prove that the function f(n) = f1 + f3 + … + f2n-1 is equal to f2n whenever n is a positive integer.

3. Problems • 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.

4. 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?

5. Problems • 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 x7y12 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).

6. The Fibonacci numbers, f0, f1, f2, … are defined as f0 = 0, f1 = 1, and fn = fn-1 + fn-2 , for n > 1. Prove that the function (n) = f1 + f3 + … + f2n-1 is equal to f2n whenever n is a positive integer. Basis Step If n = 1, then (1) = f2*1 - 1 = f1 = 1 = f2.

7. The Fibonacci numbers, f0, f1, f2, … are defined as f0 = 0, f1 = 1, and fn = fn-1 + fn-2 , for n > 1. Prove that the function (n) = f1 + f3 + … + f2n-1 is equal to f2n whenever n is a positive integer. Inductive Step Assume (k) = f1 + f3 + … + f2k-1 = f2k for k  n . We must show that this implies that (n+1) = f2(n+1) (n+1) = f1 + f3 + … + f2n-1 + f2n+1 (n+1) = (n) + f2n+1 (n+1) = f2n + f2n+1 = f2n+2 = f2(n+1)

8. 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?

9. 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.

10. 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.

11. 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

12. What is the coefficient of x12y7 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.