discrete final exam i group january solutions
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8/2/2019 Discrete Final Exam I Group January Solutions
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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.
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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
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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
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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 .
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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.
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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
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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 .
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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
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