csc 2510 test 3
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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!. 1.a. - PowerPoint PPT PresentationTRANSCRIPT
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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 suitsFind: max(suit1 + suit1 + suit1 + suit1) ≤ 3 + 1
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1.a.
max(2 + 2 + 2 + 2) ≤ 3
Answer: Sum rule (Pigeon hole accepted also)
max(2 + 2 + 2 + 2) ≤ 3 + 1 = 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 28 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 * 28 = 64 stringsBut 1/2 of the 1/4 * 28 for 32 strings also start with 1 so they must be subtracted out again. 128 + 64 - 32 = 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 choiceslength = n (or repeated n times)bits are indistinguishableeach choice is either a 0 or a 1
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1.c.
2n
(two choices for each bit - choicestimes)
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1.d.
How many ways can you solve a task if the task can be done in one of n1 ways or in one of n2 ways?
Given: n1 ways + n2 ways
Answer: Sum rule
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2.a.Convert the octal value 7016 to binary.
70168 = 111 000 001 110
Convert the octal digits to binary in groups of three using the table.
= 1110000011102
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2.b.Convert the octal value 7016 to hex.
70168 = 111 000 001 110
Convert the octal digits to binary in groups of three using the table.1110000011102
Then regroup into groups of 4 start from right: 1110 0000 11102
Then convert to hex using table (you should learn the conversion: 1110 0000 11102 = E0E16
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2.c.
Convert the octal value 7016 to decimal.
70168 = 7 * 83 + 0 *82 + 1 *81 + 6 *80
= 7 * 512 + 0 + 8 + 6 = 3584 + 8 + 6
= 359810
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2.d.Convert decimal 17861 to hex.
17861/16 = 1116 mod 5 1116/16 = 69 mod 12 = C
69/16 = 4 mod 5 4/16 = 0 mod 4
= 45C516
(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 = 33148
(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:
nr
(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:
nr
(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 aboveC(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 := ay := bwhile y 0 r := x mod y x := y y := rreturn x{gcd(a, b) is x}
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4.a.or
procedure gcd(a, b: positive integers)x := ay := bif y = 0 return xreturn gcd(y, x mod y)
orprocedure gcd(a, b: positive integers)if b = 0 return areturn gcd(b, a mod b)
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4.b.
Showing the steps, find gcd(91, 287):
287/91 = 91 * 3 + 14 91/14 = 14 * 6 + 7 14/7 = 7 * 2 + 0 (remainder of 0 indicates we are finished)
= 7
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5.
Let P(n) be the statement that 12 + 32 + 52+ ... + (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: 12 + 32 + 52 + ... + (2k + 1)2 = (k + 1)(2k + 1)(2k + 3)/3
Induction Step: Show that
12 + 32 + 52+ ... + (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:
12 + 32 + 52 + ... + (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) 2k2 + 3k + 1 + 6k + 9 = 2k2 + 9k + 10 2k2 + 9k + 10 = 2k2 + 9k + 10 1 = 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 ni=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? 2n
Same question as 3.d and 3.f.
b. How many bit strings of length 16 are there? 216
(two choices for each bit - choicestimes)
• What is the formula that you used? 2n
(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 28 plus 1/4 of 28 minus 1/4 of 1/2 of 22)(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 28 =) 128 + (1/4 of 28 = ) 64 - (1/4 of 1/2 of 26 = ) 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)
410 (10 questions with four choices each, - 410, ntimes or waysquestions)
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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)
510 (10 questions with five choices each, - 510)
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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?
n1 = 52, r1 = 5 (if in answer, accept)
n2 = 47, r2 = 5,
n3 = 42, r3= 5,
n4 = 37, r4 = 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) answersIf 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!) or52! / (13! 13! 13! 13!)
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12.bFor each hand what is the value of n? Of r?
n1 = 52, r1 =13,
n2 = 39, r2 =13,
n3 = 26, r3 =13,
n4 = 13, r4 =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.