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CSC 2510 Test 3 1. Give the names of the formula or rule that provides the answer for each of the following problems. Do not try to solve the problems!

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1.a. How many cards must be selected from a standard deck of 52 cards to guarantee that at least three cards of the same suit are chosen? Givens: Standard deck = 4 suits Find: max(suit 1 + suit 1 + suit 1 + suit 1 ) 3 + 1

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1.a. max( ) 3 Answer: Sum rule (Pigeon hole accepted also) max( ) = 9 (You can chose 2 (indistinguishable) from each group before the next card will ensure that you have three of a suit)

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1.b. How many bit strings of length eight either start with a 1 bit or end with the two bits 10? There are 2 8 strings = 256 possible strings. 1 / 2 of the strings begin with 1 for 128 strings 1 / 4 of the strings end in 10 for 1 / 4 * 2 8 = 64 strings But 1 / 2 of the 1 / 4 * 2 8 for 32 strings also start with 1 so they must be subtracted out again = 160.

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1.b. Subtraction rule.

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1.c. How many bit strings of length n are there? Given:bit strings = 2 choices length = n (or repeated n times) bits are indistinguishable each choice is either a 0 or a 1

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1.c. 2 n (two choices for each bit - choices times )

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1.d. How many ways can you solve a task if the task can be done in one of n 1 ways or in one of n 2 ways? Given: n 1 ways + n 2 ways Answer: Sum rule

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2.a. Convert the octal value 7016 to binary = Convert the octal digits to binary in groups of three using the table. =

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2.b. Convert the octal value 7016 to hex = Convert the octal digits to binary in groups of three using the table Then regroup into groups of 4 start from right: Then convert to hex using table (you should learn the conversion: = E0E 16

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2.c. Convert the octal value 7016 to decimal = 7 * * * *8 0 = 7 * = =

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2.d. Convert decimal to hex /16 = 1116 mod /16 = 69 mod 12 = C 69/16 = 4 mod 5 4/16 = 0 mod 4 = 45C5 16 (remainder of 0 indicates we are finished)

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2.e. Convert decimal 1740 to octal. 1740/8 = 217 mod 4 217/8 = 27 mod 1 27/8 = 3 mod 3 3/8 = 0 mod 3 = (remainder of 0 indicates we are finished)

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3. What formula would you use in each of the following cases? Example: permutation: P(n) = n!.

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3.a. A procedure has n1 ways to do task 1 and n2 ways to do task 2: n1 * n2 or product rule

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3.b. A task may use one of group n1 or one of n2 to do a task: n1 + n2 or sum rule (no overlap between groups – all tasks are independent)

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3.c. An ordered arrangement of r elements: P(n,r) or n!/(n - r)! or n(n-1)(n-2)...(n-r+1) (r ordered elements out of n)

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3.d. The number of r-permutations of a set of n objects with repetition: n r (with repetition means that each time you get to choice n items – the amount you can choice never gets smaller than n so that you choice n r times)

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3.e. An unordered selection of r elements from a set with n distinct elements: C(n, r) or n!/(r! * (n-r)!) (unordered is a combination and divides r! back out – (the duplicates))

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3.f. The number of r-permutations of a set of n objects with repetition: n r (Same question as 3.d.)

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3.g. The number of ways to select r items from n indistinguishable objects: (Think cookies). (the size of r may be greater than, less than, or equal to the size of n) C(n + r -1, r) or C(n + r - 1, n -1) or factorial expansion of the above C(n + r -1, r) = C(n + r - 1, n -1) because (n+ r-1)/((n+r-1-r)!*r!) = (n+ r-1)/((n-1+(r-r))!*r!) = (n+ r-1)/((n-1)!*r!)

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3.h. The number of ways to select r items from n indistinguishable objects and do it until less than r items remains: C(n,r)C(n-r,r)C(n-r-r,r) * (etc.) (Any expression of same.)

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4.a. Give the Euclidean Algorithm for greatest common divisor: procedure gcd(a, b: positive integers) x := a y := b while y 0 r := x mod y x := y y := r return x {gcd(a, b) is x}

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4.a. or procedure gcd(a, b: positive integers) x := a y := b if y = 0 return x return gcd(y, x mod y) or procedure gcd(a, b: positive integers) if b = 0 return a return gcd(b, a mod b)

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4.b. Showing the steps, find gcd(91, 287): 287/91 = 91 * /14 = 14 * /7 = 7 * (remainder of 0 indicates we are finished) = 7

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5. Let P(n) be the statement that (2n + 1) 2 = (n + 1)(2n + 1)(2n + 3)/3 whenever n is a nonnegative integer. Be sure to use the formal proof that includes Basis Step.

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5. Basis step: Show that P(0) is true. Let n = 0. Then (2(0) + 1) 2 = ((0) + 1)(2(0) + 1) (2(0) + 3)/3 1 = (1)(1)(3)/3 = 1. Since both sides equal 1, P(0) must be true Induction Hypotheses: (2k + 1) 2 = (k + 1)(2k + 1)(2k + 3)/3 Induction Step: Show that (2k + 1) 2 + (2(k+1) + 1) 2 = ((k + 1) + 1)(2(k + 1) + 1)(2(k + 1) + 3)/3 is true.

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5. Proof: (2k + 1) 2 + (2(k+1) + 1) 2 = ((k + 1) + 1)(2(k + 1) + 1)(2(k + 1) + 3)/3 (k + 1)(2k + 1)(2k + 3)/3 + (2(k+1) + 1) 2 = ((k + 1) + 1)(2(k + 1) + 1)(2(k + 1) + 3)/3 (k + 1)(2k + 1)(2k + 3)/3 + 3/3 * (2k + 3) 2 = ((k + 2)(2k + 3)(2k + 5)/3 ((k + 1)(2k + 1)(2k + 3) + 3(2k + 3) 2 )/3 = ((k + 2)(2k + 3)(2k + 5)/3 (k + 1)(2k + 1)(2k + 3) + 3(2k + 3) 2 = ((k + 2)(2k + 3)(2k + 5) (k + 1)(2k + 1) + 3(2k + 3) = ((k + 2)(2k + 5) 2k 2 + 3k k + 9 = 2k 2 + 9k k 2 + 9k + 10 = 2k 2 + 9k = 1 We have shown that the induction hypotheses is true by showing that the left side equals the right side for k + 1.

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6.a. Give a recursive algorithm (in the pseudo format specified in appendix A3) for computing: n i=0 i. procedure series(n; integer) If n = 0 return 0 return n + series (n-1) {returns the sum of the first n values}

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6.b. Show the value of n i=0 i generated at each step from 0 to n = 4. means sum, so each value is added to the previous ( means product, so each value is multiplied to the previous)

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7. Answer the following questions. a.How many bit strings of length n are there? 2 n Same question as 3.d and 3.f. b. How many bit strings of length 16 are there? 2 16 (two choices for each bit - choices times ) c.What is the formula that you used? 2 n (Same answer as a.)

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8.a. What rule would you use to compute how many bit strings of length eight either start with a 1 bit or end with the two bits 10? Subtraction rule (1/2 of 2 8 plus 1/4 of 2 8 minus 1/4 of 1/2 of 2 2 ) (areas common to both need to be subtracted out)

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8.b. How many bit strings of length eight either start with a 1 bit or end with the two bits 10? (1/2 of 2 8 =) (1/4 of 2 8 = ) 64 - (1/4 of 1/2 of 2 6 = ) 32 = 160

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9.a. A multiple-choice test contains 10 questions. There are 4 possible answers for each question. In how many ways can a student answer every question (one answer per question)? (4 ways per question, 10 questions) 4 10 (10 questions with four choices each, , n times or ways questions )

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9.b. A multiple-choice test contains 10 questions. There are 4 possible answers for each question. In how many ways can a student answer the question on the test if the student can leave answers blank? (5 ways per question, 10 questions) 5 10 (10 questions with five choices each, )

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10. Given a set with n items; a. What formula would you chose to select r items in order? P(n, r) or n!/(n-r)! b. How many possibilities would you have with n = 20 and r = 3. P(20, 3) or 20!/(20-3)! or 20*19*18 c. What is that formula called? r-permutations

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11. Show formula used reducing factors as much as possible but not necessary to multiply and divide once reduced.

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11.a. How many ways are there to distribute hands of 5 cards to each of four players from the standard deck of 52 cards? (Poker hands) C(52,5)C(47,5)C(42,5)C(37,5) or 52!/((52-5)!5!) * 47!/((47-5)!5!) * 42!/((42-5)!5!) * 37!/((37-5)!5!) = 52! / (47! * 5!) * 47! / (42! * 5!) * 42! / (37! * 5!) * 37! / (32! * 5!) = (52! / (5! * 5! * 5! * 32! * 5! ) = (52! / (5! 5! 5! 5! 32!)

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11.b. For each hand what is the value of (the first n) n? Of r? n 1 = 52, r 1 = 5 (if in answer, accept) n 2 = 47, r 2 = 5, n 3 = 42, r 3 = 5, n 4 = 37, r 4 = 5.

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11.c. Which formula should be used? C(52,5)C(47,5)C(42,5)C(37,5) or accept the following partial (incorrect) answers If C(52,5) or 52!/((52-5)! * 5!) or 52! / (47! * 5!) is in answer, accept answer.

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12. Show formula used reducing factors as much as possible but not necessary to multiply and divide once reduced.

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12.a. How many ways are there to distribute hands of 13 cards to each of four players from the standard deck of 52 cards? (Bridge hands) C(52,13)C(39,13)C(26,13)C(13,13) or 52!/((52-13)! 13!) * 39!/((39-13)! 13!) * 26!/((26-13)! 13!) * 13!/((13-13)!13!) = 52! / (39! 13!) * 39! / (26! 13!) * 26! / (13! 13!) * 13! / (0! 13!) = 52! / (1! 13!) * 1! / 1! 13!) * 1! / (1! 13!) * 1! / (0! 13!) or 52! / (13! 13! 13! 13!)

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12.b For each hand what is the value of n? Of r? n 1 = 52, r 1 =13, n 2 = 39, r 2 =13, n 3 = 26, r 3 =13, n 4 = 13, r 4 =13 If C(52,13) or 52!/((52-13)! * 13!) is in answer, accept answer.

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12.c. Which formula should be used? r-combinations or C(52,13)C(39,13)C(26,13)C(13,13) or If C(52,13) or 52!/((52-13)! * 13!) is in answer, accept answer.

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