Discrete Final Exam I Group January Solutions

8/2/2019 Discrete Final Exam I Group January Solutions

1/8

UNIVERSITY FOR INFORMATION SCIENCE AND TECHNOLOGY

Ohrid University, Macedonia

Discrete mathematics

- Final exam27.01.2012

I group

1. Let , , , 1,2,3, , , , 2,3, , ,A a b c B a b c C a b c . Find all sets (if any) M , such that:) A M M B ,

)A M M C ,

) C M M A .

(10)

Solution:

a)

, , ,1 , , , ,2 , , , ,3 ,

, , ,1,2 , , , ,1,3 , , , ,2,3

M a b c M a b c M a b c

M a b c M a b c M a b c

b) , , , 2 , , , ,3M a b c M a b c

c) Since C M M A , then this statement should be true C A . But C A , thereforesuch set M doesnt exist.


2/8

2. Which of the following compound propositions are tautologies or contradiction:a) p p q q b) p q p r p q r

i) by finding their truth table, ii) using the logical laws.

(15)Solution:

a) i)p q p q p q p q q p p q q

T T F F T F T

T F F T T T T

F T T F T F F

F F T T F F F

From the truth table above this proposition p p q q is not tautology and not

contradiction.

ii)

p p q q p p q q p p q q

p p q p q p p q p q

T q p q T p q p q

b) i)

p q r p q q r p r p q r p r p q r

p q

p r p q r

T T T T T T T T T

T T F T F F F T T

T F T F F T F F T

T F F F F F F T TF T T T T T T T T

F T F T F T T T T

F F T T F T T T T

F F F F F T T T T


3/8

From the truth table above this proposition p q p r p q r is

tautology.

ii)

p q p r p q r p q p r p q r

p q p r p q r p q p r p q r

p q p r p q r

p q p r p q r

p q p p r p q r

p q T r p q r

p q r p q r p q r p q r p r

p q r p q r p

r p q r p q T

p q r p q p r p q q r p q

p r p q q r p q T T T

3. Let 2,4,6,8,10A . Find the truth sets of the following propositional functions) : 2 7p x x x on A

b) : 2q x x x b onA

c) : 2,4,6 2,8l x x x on A

(10)

Solution:

a) pT ,b)

2

qT

c) 2,8,10qT


4/8

4. a) Prove that 4 2 549 5 3 53 2n n .b) Prove by induction that: 2 2 2 2 11 2 3 ... 1 2 1 ,

6n n n n n .

(15)

Solution:

a)Since

2

3

4

(5 5

2 2 mod49

2 9 mod 49

2 8 mod 49

2 16 mod49

2 32 mod 49 2 32 mod 49n n n

and

2

3

(4

4 4 2 4 2

3 3 mod49

3 9 mod 49

3 27 mod49

3 32 mod49

3 32 mod 49 3 3 3 9 32 mod 49

n

n n n n n

Is true that

4 2 55 3 53 2 5 9 32 53 32 mod 49

45 32 53 32 mod49

45 53 32 mod 49 98 32 mod 49

0 mod49

n n n n

n n

n n

Therefore 4 2 549 5 3 53 2n n .


5/8

b)1) 1n , 2

11 1, 1 1 1 2 1 1 1

6L R , so proposition is true for n=1,

2)

Let assume that the proposition is true for n=k, therefore

2 2 2 21

1 2 3 ... 1 2 16

k k k k ,

3) Let n=k+1,

2 2 22 2 2 2 2 2 2 2

22

2

11 2 3 ... 1 1 2 3 ... 1 1 2 1 1

6

1 2 6 61 2 1 6 11 2 1 6 1

6 6 6

31 2 2

1 2 7 6 1 2 3 22

6 6 6

11 2 2 1 1

6

k k k k k k k k

k k k k k k k k k k k k

k k kk k k k k k

k k k

Since the proposition is true for n=k+1, then the proposition is true for all positive integers.

5. A class has 10 male students and 10 female students. Find the number of ways the class canelect:

(a) two class representatives;

(b) 3 class representatives, one male and two females;

(c) a class president and vice president.

(10)

Solutions:

(a)

2020! 20 19 18!20,2 190

2 2! 20 2 ! 2!18!n C

ways.

b)

10 10 10! 10 9 8!10 10 450

1 2 2! 10 2 ! 2!8!n

ways.

(c)

20! 20!20,2 20 19 380

20 2 ! 18!n P

ways.


6/8

6. Find the number of ways 6 large books, 5 medium-size books, and 3 small books can beplaced on a shelf where:

(a) there are no restrictions;

(b) all books of the same size are together;

(b) the large books are together.

(10)

Solutions:

(a) 14 14!n P ways.

b)

number of ways placing

5 medium - size books

number of ways placing 3 small books =

number of number of ways placing

ways placing the 6 large books

books in groups by their size

3! 6 5 3 3! 6!5!3! waysn P P P

.

(c)

number of number of ways placing

ways placing the 6 large books

books considering

large books as a block

9! 6 9! 6! waysn P


7/8

7. For the following recurrence relation1 2

5 6n n n

a a a and initial conditions 0 10, 1a a ,

find:

(i)general solution; (ii) unique solution with the given initial conditions:(15)

Solution:

(i)The general soluition is obtained by first finding its characteristic polynomial x and itsroots and .

Since, 1 2 1 25 6 5 6 0n n n n n na a a a a a , its characteristic polynomial is:

2 5 6x x x with roots:

2

1 2 1 2

5 5 4 6 5 7 5 7 5 7, 6, 1

2 2 2 2x x x

Since the roots are distinct, the general solution is:

1 26 1nn

na C C , 0n

(ii) For the given initial conditions0 1

0, 1a a , the unique solution follows:

00

1 2 1 2

11

1 2 1 2

1 2

1 2

1 1 2 1

0 6 1 0

1 6 1 1 6

0

6 1

1 17 1 ,

7 7

C C C C

C C C C

C C

C C

C C C C

Therefore, 11 1 1

6 1 6 17 7 7

n nn n

na

, 0n .


8/8

8. (a) Solve the following equation for the real numbers, x and y. 2 3 2x y x y i i .

(b)Evaluate

30 25

55

22 41 1

1 2 2

i iii i

.

(15)

Solution:

(a)Since 2 3 2x y x y i i and from the equality of two complex numbers is true that:2 2 / 3

3 1

x y

x y

6 3 6

3 1

x y

x y

7 7

2 2

x

y x

1

2 1 2 0

x

y

(b) Since

15 1530 2 15 152 2 15 4 3 3 15 3 15 151 1 1 2 1 2 1 2 2 2 2 2i i i i i i i i i

11 1122 2 11 112 2 11 4 2 3 11 3 111 1 1 2 1 2 1 2 2 2 2i i i i i i i i i

12 1225 2 12 122 2 12 4 3 121 1 1 1 2 1 1 2 1 1 2 1 2 1 2 1i i i i i i i i i i i i i

22 224 2 2 22 2 2 4 4 6 2 62 2 2 2 2 1 2 1 2 2 1 2 1 2 2 2 2i i i i i i i i

Then

30 25 1215 4 455 4 13 3 3 6 6

22 4 11 6

4 6 6

1 1 2 12 2 22 1 2 1

2 21 2 2

2 2 2 16 64 64 64 81

i i i ii i i i i i

i i i ii i

i i i i i i i

Discrete Final Exam I Group January Solutions

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