CMSC 250 Spring 2016: Final Exam Review Questions Key

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1 CMSC 250 Spring 2016: Final Exam Review Questions Key 1 Description and disclaimer Use these questions to help review for the upcoming final exam. Note, these questions are not necessarily representative of the particular questions that may appear on an actual exam, nor are they representative of the level of difficulty. These are provided as a study-aide. These answers are provided for review purposes; most questions are taken verbatim from your textbook, as are many of the solutions. 2 Questions 1. Negate the following statements; make sure that the negation symbols immediately precede the predicates: (a) z y [P (x, y)] z y [ [P (x, y)] (b) x y [P (x, y) x y[q(x, y)]] x y P (x, y) x y Q(x, y) (c) x y [Q(x, y) Q(y, x)] x y[ Q(x, y) Q(y, x)] (d) y x z [(T (x, y, z) Q(x, y)] y x z[ T (x, y, z) Q(x, y)] 1

2 2. Use Strong Induction to show that the sum of an odd number of odd numbers is odd. (Hint: Rewrite the statement as an implication: if n is odd, then the sum of n odd integers is odd. ) (a) Basis: When n = 1, the sum contains only one integer, and it is odd, so the sum is odd. (b) Inductive Hypothesis: Assume true for less than n: If k < n is odd, then sum of k odd integers is odd. (c) Inductive Step: Write the sum as x 1 + x x n = (x 1 + x x n 2 ) + x n 1 + x n By the IH, x 1, x 2,..., x n 2 is odd (since it is the sum of n 2 odd integers, and n 2 is also odd), so it can be written as 2a + 1 where a is an integer. Both x n 1 and x n are odd, so they can be written as 2b + 1 and 2c + 1, for integers b and c. This gives which is odd. = (x 1 + x x n 2 ) + x n 1 + x n = (2a + 1) + (2b + 1) + (2c + 1) = 2(a + b + c + 1) + 1, Page 2

3 3. For each of the following, give a function that satisfies the requirements specified. Your function may be in parts, but it should account for each element in its domain. You may, however, restrict to codomain in order to construct your function. (a) A function f : Z 0 Z that is injective (one-to-one), but is not surjective (onto). This is pretty easy: let f : Z 0 Z be defined by the rule f(x) = x 2. (b) A function f defined over the integers, such that f is not injective (one-to-one) but is surjective (onto): Consider f : Z {0, 1}, where, { 0 x is even f(x) = 1 x is odd This maps every integer to either 0 or 1, and so it is surjective over the codomain {0, 1}, but is not injective. Page 3

4 4. These next series of questions assume that we are using the normal English alphabet that consists of 26 characters, ordered a through z. Do NOT attempt to compute numeric solutions to these questions; instead, leave your answers as exponents, factorials, and/or in terms of P (n, r) or C(n, r). (a) Allowing for repeated letters, how many words can we form of length 8? 26 8 (b) Allowing for repeated letters, find the number of words of length 8 that begin with T and end with T: 26 6 (c) Allowing for repeated letters, find the number of words of length 8 that have exactly one B: (d) Allowing for repeated letters, find the number of words of length 8 that have at least one C: Page 4

5 5. More counting questions, these dealing with subsets of the set S = {1, 2, 3,..., 10}. (a) Find the number of subsets that contain both 5 and 6: 2 8 (b) Find the number of subsets that contain exactly three elements: C(10, 3) (c) Find the number of subsets that contain exactly three elements, one of which is 3: C(9, 2) (d) Find the number of subsets that contain exactly four elements, the sum of which is odd: 2 C(5, 3) C(5, 1) Page 5

6 6. And still more counting questions, these involving combinations, etc. (a) A club as twenty-five members. How many ways can we choose a President, Vice-President, Treasurer, and Secretary? P (25, 4) = 25! = 303, 600. (25 4)! 7. Use the Pigeonhole Principle to answer the following question: A professor gives a multiple-choice quiz that has ten questions, each with four possible responses, a,b,c,d. What is the minimum number of students that must be in the professor s class in order to guarantee that at least three answer sheets must be identical? (Assume that no answers are left blank.) The pigeonholes are the answer sheets and we need to use the generalized pigeonhole principle to determine the number of pigeons needed for at least one pigeonhole to contain three pigeons. Because we have 4 10 possible answer sheets, therefore is the minimum number of students required to guarantee that three answer sheets are the same. 8. Use the Pigeonhole Principle to show that given 5 points inside of an equilateral triangle of side length 2, then at least two of the points are within 1 unit distance from each other. Draw a smaller equilateral triangle inside the first by connecting the midpoints on each of its sides. This divides the original triangle into four equilateral triangles of side length 2. Because there are 5 points, by the PHP, some two point must lie in the same smaller triangle. But, then these two points are in an equilateral triangle of side length 1, so they are within 1 unit distance of each other. Page 6