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# CSC 2510 Test 3 - PowerPoint PPT Presentation

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.

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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.

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(suit1 + suit1 + suit1 + suit1) ≤ 3 + 1

### 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)

### 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 strings

But 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.

### 1.b.

Subtraction rule.

### 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

### 1.c.

2n

(two choices for each bit - choicestimes)

### 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: n1ways +n2 ways

### 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

### 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

### 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

### 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)

### 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)

### 3.

What formula would you use in each of the following cases?

Example: permutation: P(n) = n!.

### 3.a.

A procedure has n1 ways to do task 1 andn2 ways to do task 2:

n1*n2or product rule

### 3.b.

A task may use one of group n1or one of n2 to do a task:

n1+n2or sum rule

(no overlap between groups – all tasks are independent)

### 3.c.

An ordered arrangement of r elements:

P(n,r) orn!/(n - r)! or n(n-1)(n-2)...(n-r+1)

(r ordered elements out of n)

### 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 nr times)

### 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))

### 3.f.

The number of r-permutations of a set of n objects with repetition:

nr

(Same question as 3.d.)

### 3.g.

The number of ways to select r items from n indistinguishable objects:

(thesize of r may be greater than, less than, or equal to the size of n)

C(n + r -1, r)

orC(n + r - 1, n -1)

orfactorial 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!)

### 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.)

### 4.a.

Give the Euclidean Algorithm for greatest common divisor:

proceduregcd(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}

### 4.a.

or

proceduregcd(a, b: positive integers)

x := a

y := b

if y = 0

return x

return gcd(y, x mod y)

or

proceduregcd(a, b: positive integers)

if b = 0

return a

return gcd(b, a modb)

### 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

### 5.

Let P(n) be the statement that

12 + 32 + 52+ ... + (2n + 1)2= (n + 1)(2n + 1)(2n + 3)/3whenever n is a nonnegative integer.

Be sure to use the formal proof that includes Basis Step.

### 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.

### 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.

### 6.a.

Give a recursive algorithm (in the pseudo format specified in appendix A3)

for computing:ni=0 i.

procedureseries(n; integer)

If n = 0 return 0

return n + series (n-1)

{returns the sum of the first n values}

### 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)

### 7. Answer the following questions.

• 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

### 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 28plus 1/4 of 28minus 1/4 of 1/2 of 22)

(areas common to both need to be subtracted out)

### 8.b.

How many bit strings of length eight either start with a 1 bit orend with the two bits 10?

(1/2 of 28=) 128 + (1/4 of 28 = ) 64- (1/4 of 1/2 of 26 = ) 32 = 160

### 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,ntimesor waysquestions)

### 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)

### 10.

Given a set with n items;

a. What formula would you chose to select ritems 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) or20!/(20-3)!or 20*19*18

c. What is that formula called?

r-permutations

### 11.

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

### 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!)

### 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.

### 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!)

### 12.

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

### 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!)

### 12.b

For 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

### 12.c.

Which formula should be used?

r-combinations

or

C(52,13)C(39,13)C(26,13)C(13,13)

or