Category: Discrete Mathematics Objective Questions

Discrete Mathematics Multiple Choice Questions on “Number Theory – Least Common Multiples”.

1. A Least Common Multiple of a, b is defined as __________
a) It is the smallest integer divisible by both a and b
b) It is the greatest integer divisible by both a and b
c) It is the sum of the number a and b
d) None of the mentioned

Answer: a
Clarification: Definition of LCM(a, b)-smallest multiple of a and b.

2. The LCM of two number 1, b(integer) are _________
a) b + 2
b) 1
c) b
d) None of the mentioned

Answer: c
Clarification: Since b is the smallest integer divisible by 1 and b.

3. If a, b are integers such that a > b then lcm(a, b) lies in _________
a) a>lcm(a, b)>b
b) a>b>lcm(a, b)
c) lcm(a, b)>=a>b
d) none of the mentioned

Answer: c
Clarification: LCM of number is either equal to the biggest number or greater than all.

4. LCM of 6, 10 is?
a) 60
b) 30
c) 10
d) 6

Answer: b
Clarification: Since 30 is the smallest integer divisible by 6 and 10.

5. The product of two numbers are 12 and their Greatest common divisor is 2 then LCM is?
a) 12
b) 2
c) 6
d) None of the mentioned

Answer: c
Clarification: The lcm of two number a and b is given by
lcm(a, b) = ab/(GCD(a, b)).

6. If LCM of two number is 14 and GCD is 1 then the product of two numbers is?
a) 14
b) 15
c) 7
d) 49

Answer: a
Clarification: The lcm of two number a and b is given by
lcm(a, b) = ab/(GCD(a, b)), this implies ab = lcm(a, b) * gcd(a, b).

7. If a number is 2² x 3¹ x 5⁰ and b is 2¹ x 3¹ x 5¹ then lcm of a, b is?
a) 2² x 3¹ x 5¹
b) 2² x 3² x 5²
c) 2³ x 3¹ x 5⁰
d) 2² x 3² x 5⁰

Answer: a
Clarification: Lcm is the product of sets having highest exponent value among a and b.

8. State whether the given statement is True or False.
LCM (a, b, c, d) = LCM(a,(LCM(b,(LCM(c, d)))).
a) True
b) False

Answer: a
Clarification: LCM function can be reursively defined.

9. LCM(a, b) is equals to _________
a) ab/(GCD(a, b))
b) (a+b)/(GCD(a, b))
c) (GCD(a, b))/ab
d) none of the mentioned

Answer: a
Clarification: ab = lcm(a, b)*gcd(a, b), which implies
LCM(a,b) = ab/(GCD(a,b)).

10. The lcm of two prime numbers a and b is _________
a) ^a⁄_b
b) ab
c) a + b
d) 1

Answer: b
Clarification: LCM(a, b) = ab/(GCD(a, b)), Since (GCD(a, b)) = 1 therfore LCM(a, b) = ab.

Discrete Mathematics Multiple Choice Questions on “Counting – Linear Permutation”.

1. How many substrings (of all lengths inclusive) can be formed from a character string of length 8? (Assume all characters to be distinct)
a) 14
b) 21
c) 54
d) 37

Answer: d
Clarification: Let’s consider the given string is CLEAN, so set of string of length 1 = {C,L,E,A,N} ; cardinality of set = 5 set of string of length 2 = {CL,EE,EA,NN}, set of string of length 3 = {CLE,LEE,EAN}, set of strings of length 4 = {CLEN,LEAN}, set of strings of length 5 = {CLEAN} and set of string of length 0 = {} and we cannot have any substring of length 6 as given string has only 5 length. So total no of substrings are possible = 0 length substring + 1 length substring + 2 length substrings +3 length substrings + 4 length substrings + 5 length substrings = 1 + 5 + 4 + 3 + 2 + 1 = 16 means for 1 length string to n length substrings it will sum of the n natural no from 1 to n.
so 1+2+3+…+n = (frac{n(n+1)}{2}) so total no substrings possible = 0 length strings + (frac{n(n+1)}{2}) = 1+ ([frac{n(n+1)}{2}]) so total no of substrings possible in n length string (All length inclusive )= 1 + ([frac{n(n+1)}{2}] = frac{8(8+1)}{2}) = 37.

2. The number of diagonals can be drawn in a hexagon is ______
a) 9
b) 32
c) 16
d) 21

Answer: a
Clarification: A hexagon has 6 sides. We obtain the diagonals by joining the vertices in pairs.
Total number of sides and diagonals = ⁶C₂ = (frac{6 * 5}{2 * 1}) = 5×3 = 15. This includes its 6 sides also. So, Diagonals = 15 – 6 = 9. Hence, the number of diagonals is 9.

3. The number of binary strings of 17 zeros and 8 ones in which no two ones are adjacent is ___________
a) 43758
b) 24310
c) 32654
d) 29803

Answer: a
Clarification: First place 17 zeroes side by side _ 0 _ 0 _ 0 … 0 _ and 8 1’s can be placed in any of the (17+1) available gaps hence the number of ways = ⁿ⁺¹C_k = 43758.

4. How many words that can be formed with the letters of the word ‘SWIMMING’ such that the vowels do not come together? Assume that words are of with or without meaning.
a) 430
b) 623
c) 729
d) 1239

Answer: c
Clarification: The word ‘SWIMMING contains 8 letters. Of which, I occurs twice and M occurs twice. Therefore, the number of words formed by this word = (frac{8!}{2!*2!}) = 10080. In order to find the number of permutations that can be formed where the two vowels I and I come together, we group the letters that should come together and consider that group as one letter. So, the letters are S, W, M, M, N, G, (I, I). So, the number of letters are 7 the number of ways in which 7 letters can be arranged is 7! = 5040. In I and I, the number of ways in which I and I can be arranged is 2!. Hence, the total number of ways in which the letters of the ‘SWIMMING’ can be arranged such that vowels are always together are (frac{7!}{2!*2!}) = 5040 ways. The number of words in which the vowels do not come together is = (10080 – 5040) = 5040.

5. A number lock contains 6 digits. How many different zip codes can be made with the digits 0–9 if repetition of the digits is allowed upto 3 digits from the beginning and the first digit is not 0?
a) 254307
b) 453600
c) 458760
d) 972340

Answer: b
Clarification: For the first position, there are 9 possible choices (since 0 is not allowed). After that number is chosen, there are 10 possible choices (since 0 is now allowed) for the second digit, for the third digit there are 10 possible choices, 9 possible choices for the fourth digit and 8 possible choices for the fifth digit and 7 possible choices for the sixth digit. The count of number locks = 453600.

6. Let M be a sequence of 9 distinct integers sorted in ascending order. How many distinct pairs of sequences, N and O are there such that i) each are sorted in ascending order, ii) N has 5 and O has 4 elements, and iii) the result of merging N and O gives that sequence?
a) 84
b) 35
c) 194
d) 138

Answer: a
Clarification: Selecting any 3 elements from given 9 elements gives ⁹C₃ = 84 number of distinct pairs of sequences.

7. 14 different letters of alphabet are given, words with 6 letters are formed from these given letters. How many number of words are there which have at least one letter repeated?
a) 892742
b) 999988
c) 213216
d) 786730

Answer: b
Clarification: Number of words which have at least one letter replaced = total number of words – total number of words in which no letter is repeated, => 10⁶ – ¹²P₆ => 1000000 − 924 = 999988.

8. In how many ways can 10 boys be seated in a row having 28 seats such that no two friends occupy adjacent seats?
a) ¹³P₅
b) ⁹P₂₉
c) ¹⁹P₁₀
d) ¹⁵P₇

Answer: c
Clarification: First let us take the 18 unoccupied seats. They create 19 slots i.e., one on the left of each seat and one on the right of the last one. So we can place the 10 boys in any of these 19 slots that are, ¹⁹P₁₀ ways.

9. In how many ways can the letters of the word be rearranged such that the vowels always appear together?
a) (frac{(8 + 3)!}{2!})
b) (frac{6!}{2!})
c) 8!*3!
d) (frac{4!}{8!})

Answer: c
Clarification: Take AOU together and treat it like 1 entity and arrange SNFNDRY in 8! Ways. Then, the AOU can be arranged in 3! ways. So, total arrangements = 8! * 3! = 40320 * 6 = 241920.

10. How many ways can 8 prizes be given away to 7 students, if each student is eligible for all the prizes?
a) 40325
b) 40320
c) 40520
d) 40720

Answer: b
Clarification: Now the first student is eligible to receive any of the 8 available prizes (so 8 ways), the second student will receive a prize from rest 7 available prizes (so 7 ways), the third student will receive his prize from the rest 6 prizes available(so 6 ways) and so on. So total ways would be 8! = 8*7*6*5*4*3*2*1 = 40320. Hence, the 7 prizes can be distributed in 40320 ways.

Discrete Mathematics Multiple Choice Questions on “Discrete Probability – Generating Functions”.

1. What is the sequence depicted by the generating series 4 + 15x² + 10x³ + 25x⁵ + 16x⁶+⋯?
a) 10, 4, 0, 16, 25, …
b) 0, 4, 15, 10, 16, 25,…
c) 4, 0, 15, 10, 25, 16,…
d) 4, 10, 15, 25,…

Answer: c
Clarification: Consider the coefficients of each xⁿ term. So a₀=4, since the coefficient of x₀ is 4 (x₀=1 so this is the constant term). Since 15 is the coefficient of x², so 15 is the term a₂ of the sequence. To find a₁ check the coefficient of x₁ which in this case is 0. So a₁=0. Continuing with these we have a₂=15, a₃=10, a₄=25, and a₅=16. So we have the sequence 4, 0, 15, 10, 25, 16,…

2. What is the generating function for the sequence 1, 6, 16, 216,….?
a) (frac{(1+6x)}{x^3})
b) (frac{1}{(1-6x)})
c) (frac{1}{(1-4x)})
d) 1-6x²

Answer: b
Clarification: For the sequence 1, 6, 36, 216,… the generating function must be (frac{1}{(1-6x}), when basic generating function: (frac{1}{1-x}).

3. What is the generating function for generating series 1, 2, 3, 4, 5,… ?
a) (frac{2}{(1-3x)})
b) (frac{1}{(1+x)})
c) (frac{1}{(1−x)^2})
d) (frac{1}{(1-x2)})

Answer: c
Clarification: Basic generating function is (frac{1}{1-x}). If we differentiate term by term in the power series, we get (1 + x + x² + x³ +⋯)′ = 1 + 2x + 3x² + 4x³ +⋯ which is the generating series for 1, 2, 3, 4,….

4. What is the generating function for the generating sequence A = 1, 9, 25, 49,…?
a) 1+(A-x²)
b) (1-A)-1/x
c) (1-A)+1/x²
d) (A-x)/x³

Answer: b
Clarification: The generating function for the sequence A. Using differencing:
A = 1 + 9x + 25x² + 49x³ + ⋯(1)
−xA = 0 + x + 9x² + 25x³ + 49x⁴ + ⋯(2)
(1−x)A = 1 + 8x + 16x² + 24x³ +⋯. Since 8x + 16x² + 24x³ + ⋯ = (1-x)A-1 ⇒ 8 + 16x + 24x² +…= (1-A)-1/x.

5. What is the recurrence relation for the sequence 1, 3, 7, 15, 31, 63,…?
a) a_n = 3a_n-1−2a_n+2
b) a_n = 3a_n-1−2a_n-2
c) a_n = 3a_n-1−2a_n-1
d) a_n = 3a_n-1−2a_n-3

Answer: b
Clarification: The recurrence relation for the sequence 1, 3, 7, 15, 31, 63,… should be a_n = 3a_n-1−2a_n-2. The solution for A: A=1/1 − 3x + 2x².

6. What is multiplication of the sequence 1, 2, 3, 4,… by the sequence 1, 3, 5, 7, 11,….?
a) 1, 5, 14, 30,…
b) 2, 8, 16, 35,…
c) 1, 4, 7, 9, 13,…
d) 4, 8, 9, 14, 28,…

Answer: a
Clarification: The first constant term is 1⋅1, next term will be 1⋅3 + 2⋅1 = 5, the next term: 1⋅5 + 2⋅3 + 3⋅1 = 14, another one: 1⋅7 + 2⋅5 + 3⋅3 + 4⋅1 = 30. The resulting sequence is 1, 5, 14, 30,…

7. What will be the sequence generated by the generating function 4x/(1-x)²?
a) 12, 16, 20, 24,…
b) 1, 3, 5, 7, 9,…
c) 0, 4, 8, 12, 16, 20,…
d) 0, 1, 1, 3, 5, 8, 13,…

Answer: c
Clarification: The sequence should be 0, 4, 8, 12, 16, 20,…for the generating function 4x/(1-x)², when basic generating function: 1/(1-x).

8. What is the generating function for the sequence with closed formula a_n=4(7ⁿ)+6(−2)ⁿ?
a) (4/1−7x)+6!
b) (3/1−8x)
c) (4/1−7x)+(6/1+2x)
d) (6/1-2x)+8

Answer: c
Clarification: For the given sequence after evaluating the formula the generating formula will be (4/1−7x)+(6/1+2x).

9. Suppose G is the generating function for the sequence 4, 7, 10, 13, 16, 19,…, the find a generating function (in terms of G) for the sequence of differences between terms.
a) (1−x)G−4/x
b) (1−x)G−4/x³
c) (1−x)G+6/x
d) (1−x)G−x²

Answer: a
Clarification: (1−x)G = 4 + 3x + 6x² + 9x³ +⋯ which can be accepted. We can compute it like this:
3 + 6x + 9x² + ⋯ = (1−x)G−4/x.

10. Find the sequence generated by 1/1−x²−x⁴.,assume that 1, 1, 2, 3, 5, 8,… has generating function 1/1−x−x².
a) 0, 0, 1, 1, 2, 3, 5, 8,…
b) 0, 1, 2, 3, 5, 8,…
c) 1, 1, 2, 2, 4, 6, 8,…
d) 1, 4, 3, 5, 7,…

Answer: a
Clarification: Based on the given generating function, the sequence will be 0, 0, 1, 1, 2, 3, 5, 8,… which is generated by 1/1−x²−x⁴.

Discrete Mathematics Objective Questions & Answers on “Different Path in a Graph”.

1. Which algorithm efficiently calculates the single source shortest paths in a Directed Acyclic Graph?
a) topological sort
b) hash table
c) binary search
d) radix sort

Answer: a
Clarification: For Directed Acyclic graph, single source shortest distances can be calculated in O(V+E) time. For that purpose Topological Sorting can be used. Topological Sorting of any graph represents a linear ordering of the graph.

2. The _______ of a graph G consists of all vertices and edges of G.
a) edge graph
b) line graph
c) path complement graph
d) eulerian circuit

Answer: d
Clarification: we know that he Eulerian circuit in a graph G is a circuit that includes all vertices and edges of G. A graph that can have Eulerian circuit, also can have a Eulerian graph.

3. A ______ in a graph G is a circuit which consists of every vertex (except first/last vertex) of G exactly once.
a) Euler path
b) Hamiltonian path
c) Planar graph
d) Path complement graph

Answer: b
Clarification: The Eulerian path in a graph say, G is a walk from one vertex to another, that can pass through all vertices of G as well as traverses exactly once every edge of G. Therefore, an Eulerian path can not be a circuit. A Hamiltonian path is a walk that contains every vertex of the graph exactly once. Hence, a Hamiltonian path is not a circuit.

4. A walk has Closed property if ____________
a) v₀=v_k
b) v₀>=v_k
c) v < 0
d) v_k > 1

Answer: a
Clarification: A walk in a graph is said to be closed if the starting vertex is the same as the ending vertex, that is v₀=v_k, it is described as Open otherwise.

5. A trail in a graph can be described as ______________
a) a walk without repeated edges
b) a cycle with repeated edges
c) a walk with repeated edges
d) a line graph with one or more vertices

Answer: a
Clarification: Suppose in a graph G a trail could be defined as a walk with no repeated edges. Suppose a walk can be defined as efgh. There are no repeated edges so this walk is a trail.

6. Let a graph can be denoted as ncfkedn a kind of ____________
a) cycle graph
b) line graph
c) hamiltonian graph
d) path graph

Answer: a
Clarification: In the graph ncfkedn, no edges are repeated in the walk, which makes it a trail and then start and end vertex n is same making it a cycle graph.

7. Determine the edge count of a path complement graph with 14 vertices.
a) 502
b) 345
c) 78
d) 69

Answer: c
Clarification: Let, an n-path complement graph P_n’ is the graph complement of the path graph P_n. Since P_n is self-complementary, P₄’ is isomorphic to P₄. Now, P_n’ has an edge count = ¹⁄₂(n-2)(n-1). So, the required edge count is=78.

8. The sum of an n-node graph and its complement graph produces a graph called _______
a) complete graph
b) bipartite graph
c) star graph
d) path-complement graph

Answer: a
Clarification: Suppose, the complement G’ of a graph G is known as edge-complement graph which consists of with the same vertex set but whose edge set contains the edges not present in G. The graph sum G+G’ on an n-node graph G is called the complete graph say, K_n.

9. In a directed weighted graph, if the weight of every edge is decreased by 10 units, does any change occur to the shortest path in the modified graph?
a) 209
b) 65
c) 57
d) 43

Answer: c
Clarification: The shortest path will change in the modified graph. Suppose that the shortest path is of weight 21 and has 7 edges and there is another path with 4 edges and total weight 17. Now, the weight of the first shortest path is increased by 7*10 and becomes 21 + 70 and the weight of the second path is increased by 4*10 and becomes 17 + 40. So the shortest path changes to the other path with weight as 57.

10. Let G(V, E) be a directed graph where every edge has weight as either 1, 2 or 5, what is the algorithm used for the shortest path from a given source vertex to a given destination vertex to get the time complexity of O(V+E)?
a) BFS
b) DFS
c) Binary search
d) Radix sort

Answer: a
Clarification: In BFS due to the least number of edges between two vertices and so if all the edges in a graph are of same weight, then to find the shortest path BFS can be used for efficiency. So we have to split all edges of weight 5 into two edges of weight 2 each and one edge of weight 1. In the worst case, all edges are of weight 1. To split all edges, O(E) operations can be done and so the time complexity becomes which is equal to O(V+E).

Discrete Mathematics Multiple Choice Questions on “Group Axioms”.

1. __________ are called group postulates.
a) Group lemmas
b) Group theories
c) Group axioms
d) Group

Answer: c
Clarification: The group axioms are also called the group postulates. A group with an identity (that is, a monoid) in which every element has an inverse is termed as semi group.

2. A subgroup has the properties of ________
a) Closure, associative
b) Commutative, associative, closure
c) Inverse, identity, associative
d) Closure, associative, Identity, Inverse

Answer: d
Clarification: A subgroup S is a subset of a group G (denoted by S <= G) if it holds the four properties simultaneously – Closure, Associative, Identity and Inverse element.

3. If a * b = a such that a ∗ (b ∗ c) = a ∗ b = a and (a * b) * c = a * b = a then ________
a) * is associative
b) * is commutative
c) * is closure
d) * is abelian

Answer: a
Clarification: ‘∗’ can be defined by the formula a∗b = a for any a and b in S. Hence, (a ∗ b)∗c = a∗c = a and a ∗(b ∗ c)= a ∗ b = a. Therefore, ”∗” is associative. Hence (S, ∗) is a semigroup. On the contrary, * is not associative since, for example, (b•c)•c = a•c = c but b•(c•c) = b•a = b Thus (S,•) is not a semigroup.

4. The set of odd and even positive integers closed under multiplication is ________
a) a free semigroup of (M, ×)
b) a subsemigroup of (M, ×)
c) a semigroup of (M, ×)
d) a subgroup of (M, ×)

Answer: b
Clarification: Let C and D be the set of even and odd positive integers. Then, (C, ×) and (D, ×) are subsemigroups of (M, ×) since A and B are closed under multiplication. On the other hand, (A, +) is a subsemigroup of (N, +) since A is closed under addition, but (B, +) is not a subsemigroup of (N, +) since B is not closed under addition.

5. If F is a free semigroup on a set S, then the concatenation of two even words is ________
a) a semigroup of F
b) a subgroup of F
c) monoid of F
d) cyclic group of F

Answer: b
Clarification: Let F be the free semigroup on the set S = {m,n}. Let, E consist of all even words, i.e, words with even length and the concatenation of two such words is also even. Thus E is a subsemigroup of F.

6. The set of rational numbers form an abelian group under _________
a) Association
b) Closure
c) Multiplication
d) Addition

Answer: c
Clarification: The set of nonzero rational numbers form an abelian group under multiplication. The number 1 is the identity element and q/p is the multiplicative inverse of the rational number p/q.

7. Condition of semigroup homomorphism should be ____________
a) f(x * x) = f(x * y)
b) f(x) = f(y)
c) f(x) * f(y) = f(y)
d) f(x * y) = f(x) * f(y)

Answer: d
Clarification: Consider two semigroups (S,∗) and (S’,∗’). A function f: S -> S’ is called a semigroup homomorphism if f(a∗b) = f(a)∗f(b). Suppose f is also one-to-one and onto. Then f is called an isomorphism between S and S’ and S and S’ are said to be isomorphic semigroups.

8. A function f:(M,∗)→(N,×) is a homomorphism if ______
a) f(a, b) = a*b
b) f(a, b) = a/b
c) f(a, b) = f(a)+f(b)
d) f(a, b) = f(a)*f(a)

Answer: b
Clarification: The function f is a homomorphism since f(x∗y)= f(ac, bd)= (ac)/(bd) = (a/b)(c/d) = f(x)f(y).

9. A function defined by f(x)=2*x such that f(x+y)=2x+y under the group of real numbers, then ________
a) Isomorphism exists
b) Homomorphism exists
c) Heteromorphic exists
d) Association exists

Answer: b
Clarification: Let T be the group of real numbers under addition, and let T’ be the group of positive real numbers under multiplication. The mapping f: T -> T’ defined by f(a)=2*a is a homomorphism because f(a+b)=2a+b = 2a*2b = f(a)*f(b). Again f is also one-to-one and onto T and T’ are isomorphic.

10. If x * y = x + y + xy then (G, *) is _____________
a) Monoid
b) Abelian group
c) Commutative semigroup
d) Cyclic group

Answer: c
Clarification: Let x and y belongs to a group G.Here closure and associativity axiom holds simultaneously. Let e be an element in G such that x * e = x then x+e+xe=a => e(1+x)=0 => e = 0/(1+x) = 0. So, identity axiom does not exist but commutative property holds. Thus, (G,*) is a commutative semigroup.

Discrete Mathematics Multiple Choice Questions on “Set Operations”.

1. The union of the sets {1, 2, 5} and {1, 2, 6} is the set _______________
a) {1, 2, 6, 1}
b) {1, 2, 5, 6}
c) {1, 2, 1, 2}
d) {1, 5, 6, 3}

Answer: b
Clarification: The union of the sets A and B, is the set that contains those elements that are either in A or in B.

2. The intersection of the sets {1, 2, 5} and {1, 2, 6} is the set _____________
a) {1, 2}
b) {5, 6}
c) {2, 5}
d) {1, 6}

Answer: a
Clarification: The intersection of the sets A and B, is the set containing those elements that are in both A and B.

3. Two sets are called disjoint if there _____________ is the empty set.
a) Union
b) Difference
c) Intersection
d) Complement

Answer: c
Clarification: By the definition of the disjoint set.

4. Which of the following two sets are disjoint?
a) {1, 3, 5} and {1, 3, 6}
b) {1, 2, 3} and {1, 2, 3}
c) {1, 3, 5} and {2, 3, 4}
d) {1, 3, 5} and {2, 4, 6}

Answer: d
Clarification: Two sets are disjoint if the intersection of two sets is the empty set.

5. The difference of {1, 2, 3} and {1, 2, 5} is the set ____________
a) {1}
b) {5}
c) {3}
d) {2}

Answer: c
Clarification: The difference of the sets A and B denoted by A-B, is the set containing those elements that are in A not in B.

6. The complement of the set A is _____________
a) A – B
b) U – A
c) A – U
d) B – A

Answer: b
Clarification: The complement of the set A is the complement of A with respect to U.

7. The bit string for the set {2, 4, 6, 8, 10} (with universal set of natural numbers less than or equal to 10) is ____________________
a) 0101010101
b) 1010101010
c) 1010010101
d) 0010010101

Answer: a
Clarification: The bit string for the set has a one bit in second, fourth, sixth, eighth, tenth positions, and a zero elsewhere.

8. Let Ai = {i, i+1, i+2, …..}. Then set {n, n+1, n+2, n+3, …..} is the _________ of the set Ai.
a) Union
b) Intersection
c) Set Difference
d) Disjoint

Answer: b
Clarification: By the definition of the generalized intersection of the set.

9. The bit strings for the sets are 1111100000 and 1010101010. The union of these sets is ___________
a) 1010100000
b) 1010101101
c) 1111111100
d) 1111101010

Answer: d
Clarification: The bit string for the union is the bitwise OR of the bit strings.

10. The set difference of the set A with null set is __________
a) A
b) null
c) U
d) B

Answer: a
Clarification: The set difference of the set A by the null set denoted by A – {null} is A.