Computing Science 272 Solutions to Midterm Examination I Tuesday February 8, 2005


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1 Computing Science 272 Solutions to Midterm Examination I Tuesday February 8, 2005 Department of Computing Science University of Alberta Question 1. 8 = pts (a) How many 16bit strings contain exactly nine 1 s? (b) How many 16bit strings contain at least fourteen 1 s? (c) How many 16bit strings contain at least one 1? (d) How many 16bit strings contain at most one 1? Solution: (a) The number of 16bit strings containing exactly nine 1 s is just the number of ways of choosing the bits to put the 1 s into, that is, ( ) 16 = 16! 9 9! 7! = (b) The number of 16bit strings containing at least fourteen 1 s is the same as the number which contain either 14, 15 or 16 1 s, that is ( ) ( ) ( ) = = (c) The number of 16bit strings that contain at least one 1 is the total number of 16bit strings minus the number of 16bit strings that contain no 1 s, that is, ( ) = = 65, (d) The number of 16bit strings that contain at most one 1 is the same as the number which contain zero 1 s plus the number that contain one 1, that is, ( ) ( ) = = Question 2. 8 = pts A computer programming team has 14 members. (a) How many ways can a group of seven be chosen to work on a project? (b) Suppose eight team members are women and six are men. (i) How many groups of seven can be chosen that contain four women and three men? (ii) How many groups of seven can be chosen that contain at least one man? (iii) How many groups of seven can be chosen that contain at most three women? (c) Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project? (d) Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?
2 Solution: (a) The number of ways to select 7 members from 14 members is ( ) 14 = 14! 7 7! 7! = (b) Suppose that 8 team members are women and 6 team members are men. (i) From the rule of product, the number of groups of 7 that contain 4 women and 3 men is the number of ways of selecting 4 women from the 8 women times the number of ways of selecting 3 men from the 6 men, and this is ( ) 8 4 ( ) 6 3 = 8! 4! 4! (ii) The number of groups of 7 that contain no men is ( ) ( ) 8 8 = = 8, 7 1 6! = = ! 3! and the number of groups of 7 in total is ( ) 14, 7 so that the number of groups of 7 that contain at least 1 man is ( ) ( ) 14 8 = 14! 8 = = ! 7! (iii) From the rule of sum, the number of groups of 7 that contain at most 3 women is the sum of the number of groups that contain 0, 1, or 2 women, that is, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = = = (c) Let a and b be the two people who refuse to work together, and let A and B be the number of groups that can be selected from the 14 people where a is selected and where b is selected, respectively. Also, let N be the number of groups where neither a nor b work, then A = B = ( ) 12 6 = 12! = 924, and N = 6! 6! ( ) 12 = 12! 7 7! 5! = 792, so that the number of possible groups of 7 that can be formed under these circumstances is A + B + N = = (d) Let a and b be the two team members who insist on working together or not at all, and let B be the number of groups that can be selected when both of them work and let N be the number of groups that can be selected when neither of them work, then from the rule of sum, the number of groups of 7 that can be selected under these circumstances is ( ) ( ) ( ) ! B + N = + = 2 = 2 = = ! 7!
3 Question 3. 8 pts In how many different ways can twelve people, denoted by A, B,..., L be seated at a square table (as shown below), where A and B refuse to sit next to each other (even at the corner of the table)? A B C L D K E J F I H Note: As in the assignment problem, if one arrangement can be obtained from another arrangement by a rotation of 90, 180, or 270 degrees, then they are considered to be the same. Solution: There are three mutually exclusive cases to consider, depending on where A is seated. Either there is no one seated to the right of A, or there is no one seated to the left of A, or A is seated between two other people. In each case, we have 9 positions in which B can be seated, and for each of these there are 10! ways to arrange the other 10 people. By the rule of product, there are 9 10! different arrangements in each case. Finally, by the rule of sum, there are 9 10! ! ! = 27 10! different seatings where A and B are not sitting together. Question 4. 8 = pts The logical connective Nor is defined by the truth table below, it is denoted by the symbol. G p q p q Thus, p q means not (p or q). In this problem you are to show that all compound statements can be written using only this connective. (a) Find a statement logically equivalent to p using only the connective. p p p (b) Find a statement logically equivalent to p q using only the connective. p q (p p) (q q) (c) Find a statement logically equivalent to p q using only the connective. p q (p q) (p q) (d) Find a statement logically equivalent to p q using only the connective. p q ((p p) q) ((p p) q)
4 Question = 6+4 pts In a new case, Sherlock Holmes has discovered that (i) The cook or the butler was in the kitchen. (ii) The cook was in the kitchen or in the dining room. (iii) If the butler was smoking a cigar, he was not in the kitchen. (iv) If the cook was not in the dining room, the butler was not smoking a cigar. Define the following propositions: b = The butler was in the kitchen. c = The cook was in the kitchen. d = The cook was in the dining room. s = The butler was smoking a cigar. (a) Fill in the truth table below for all the statements contained in the premises (i), (ii), (iii), and (iv). b c d s b c c d s b d s (b) Which of the statements b, c, d, s, or their negations, can Holmes deduce from the facts above? You must list the reasons, that is, the rules of inference, for each of the statements in your proof. There are only three rows for which all the premises are true, namely for b = 0, c = 1, d = 0, s = 0, for b = 1, c = 0, d = 1, s = 0, and for b = 1, c = 1, d = 0, s = 0; thus, the implication s is a tautology. Therefore, if we assume the premises are true, then Modus Ponens gives s, that is, the only statement Holmes can deduce from these premises is s.
5 The theorem we want to prove is s and we do this by contradiction, we form the conjunction of all the premises with the negation of the conclusion and show that this is a contradiction; so the premises are s, and the proof of the theorem is given below. Question 6. 8 = pts Steps Rule 1. b c premise 2. c d premise 3. s b premise 4. d s premise 5. s assumption 6. b 3, 5, modus ponens 7. c 1, 6, disjunctive syllogism 8. d 2, disjunctive syllogism 9. s 4, 8, modus ponens 10. s 5, 9, contradiction Using the following predicates on the universal set of all persons, Predicate Mother(A, B) Father(A, B) Female(A) Male(A) Sister(A, B) Brother(A, B) Older(A, B) Intended Meaning A is the mother of B A is the father of B A is female A is male A is the sister of B A is the brother of B A is older than B express each of the following statements in symbolic form: (a) No person is older than that person s father. A B Father(B, A) Older(A, B) (b) Some persons have no younger brothers. A B Brother(B, A) Older(A, B) (c) If any person A has the same mother and father as another person B, and if A is female, then A is a sister of B. A B ( M F Mother(M, A) Mother(M, B) Father(F, A) Father(F, B) ) ( A B ) Female(A) Sister(A, B) (d) If any person A is the brother or sister of another person B, then A has the same mother as B, or A has the same father as B. A B A B ( Brother(A, B) Sister(A, B) ) ( MMother(M, A) Mother(M, B) ) ( F Father(F, A) Father(F, B) )
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