Examination Sheet № 2
Examination Sheet № 1
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| Propositions. Logical Operations: negation, conjunction, disjunction, implication, exclusive or, biconditional. Truth Tables. Bitwise logical operations. Tautology and a contradiction. Prove that (p → (q → r)) → ((p → q) → (p → r)) is a tautology without using truth tables. Propositional Equivalences. Tables of Logical Equivalences.
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| Recurrence relations. Solution of a recurrence relation. The tower of Hanoi. Codeword enumeration.
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| Construct the DNF, CNF and a polynomial for a proposition F (p, q, r) which is true iff (p, q, r) are from {(1, 1, 0), (1, 0, 0), (0, 0, 1), (0, 0, 0)}
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| How many ways to select 23 ordered elements from a set consisting of 54 elements are there, if repetition is not allowed?
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| Find a sequence { an: n = 0, 1, …} satisfying the recurrence relation xn + 6 = –6(xn –1 – 1) – 9 xn –2, with the initial conditions a 0 =1, a 1 = –9.
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Examination Sheet № 2
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| Truth Tables of basic logical operations. The number of all compound propositions consisting of n elementary compositions. Prove this by mathematical induction. Construct the truth table for the proposition  .
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| Generalized permutation and combinations. Permutations and combinations with repetition. Permutations of sets with indistinguishable objects, distributing objects into boxes.
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| Construct the DNF, CNF and a polynomial for a proposition F (p, q, r) which is true iff (p, q, r) are from {(1, 1, 0), (1, 0, 0), (0, 0, 1), (0, 0, 0)}
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| How many ways to select 23 ordered elements from a set consisting of 54 elements are there, if repetition is not allowed?
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| Find a sequence { an: n = 0, 1, …} satisfying the recurrence relation xn + 6 = –6(xn –1 – 1) – 9 xn –2, with the initial conditions a 0 =1, a 1 = –9.
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2015-07-14
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