Category: Discrete Mathematics Objective Questions

Discrete Mathematics Multiple Choice Questions on “Counting – Combinations”.

1. There are 6 equally spaced points A, B, C, D, E and F marked on a circle with radius R. How many convex heptagons of distinctly different areas can be drawn using these points as vertices?
a) 7! * 6
b) 7C5
c) 7!
d) same area

Answer: d
Clarification: Since all the points are equally spaced; hence the area of all the convex heptagons will be the same.

2. There are 2 twin sisters among a group of 15 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two sisters?
a) 15 *12! * 2!
b) 15! * 2!
c) ¹⁴C₂
d) 16 * 15!

Answer: a
Clarification: We know that n objects can be arranged around a circle in (frac{(n−1)!}{2}). If we consider the two sisters and the person in between the brothers as a block, then there will 12 others and this block of three people to be arranged around a circle. The number of ways of arranging 13 objects around a circle is in 12! ways. Now the sisters can be arranged on either side of the person who is in between the sisters in 2! ways. The person who sits in between the two sisters can be any of the 15 in the group and can be selected in 15 ways. Therefore, the total number of ways 15 *12! * 2!.

3. The number of words of 4 consonants and 3 vowels can be made from 15 consonants and 5 vowels, if all the letters are different is ________
a) 3! * ¹²C₅
b) ¹⁶C₄ * ⁴C₄
c) 15! * 4
d) ¹⁵C₄ * ⁵C₃ * 7!

Answer: d
Clarification: There are 4 consonants out of 15 can be selected in ¹⁵C₄ ways and 3 vowels can be selected in ⁵C₃ ways. Therefore, the total number of groups each containing 4 consonants and 3 vowels = ¹⁵C₄ * ⁴C₃. Each group contains 7 letters which can be arranged in 7! ways. Hence, required number of words = ¹⁵C₄ * ⁵C₃ * 7!.

4. How many ways are there to arrange 7 chocolate biscuits and 12 cheesecake biscuits into a row of 19 biscuits?
a) 52347
b) 50388
c) 87658
d) 24976

Answer: b
Clarification: Consider the situation as having 19 spots and filling them with 7 chocolate biscuits and 19 cheesecake biscuits. Then we just choose 7 spots for the chocolate biscuits and let the other 10 spots have cheesecake biscuits. The number of ways to do this job is ¹⁹C₇ = 50388.

5. If a, b, c, d and e are five natural numbers, then find the number of ordered sets(a, b, c, d, e) possible such that a+b+c+d+e=75.
a) ⁶⁵C₅
b) ⁵⁸C₆
c) ⁷²C₇
d) ⁷⁴C₄

Answer: d
Clarification: Let assumes that there are 75 identical balls which are to be arranged in 5 different compartments (Since a, b, c, d, e are distinguishable). If the balls are arranged in the row. We have 74 gaps where we can place a ball in each gap since we need 5 compartments we need to place only 4 balls. We can do this in ⁷⁴C₄ ways.

6. There are 15 people in a committee. How many ways are there to group these 15 people into 3, 5, and 4?
a) 846
b) 2468
c) 658
d) 1317

Answer: d
Clarification: The number of ways to choose 3 people out of 9 is ¹⁵C₃. Then, number of ways to choose 5 people out of (15-3) = 12 is ¹²C₅. Finally, the number of ways to choose 4 people out of (12-4) = 8 is ⁸C₄. Hence, by the rule of product, ¹⁵C₃ + ¹²C₅ + ⁸C₄ = 1317.

7. There are six movie parts numbered from 1 to 6. Find the number of ways in which they be arranged so that part-1 and part-3 are never together.
a) 876
b) 480
c) 654
d) 237

Answer: b
Clarification: The total number of ways in which 6 part can be arranged = 6! = 720. The total number of ways in which part-1 and part-3 are always together: = 5!*2! = 240. Therefore, the total number of arrangements, in which they are not together is = 720 − 240 = 480.

8. How many ways are there to divide 4 Indian countries and 4 China countries into 4 groups of 2 each such that at least one group must have only Indian countries?
a) 6
b) 45
c) 12
d) 76

Answer: a
Clarification: The number of ways to divide 4+4=8 countries into 4 groups of 2 each is as follows: (¹⁰C₂ * ¹⁰C₂ * ¹⁰C₂ * ¹⁰C₂)/4! = 30. Since it is required that at least one group must have only Indian countries, we need to subtract 30 from the number of possible groupings where all 4 groups have 1 Indian country and 1 China country each. This is equivalent to the number of ways to match each of the 4 Indian countries with one China country: 4! = 24. Therefore, the answer is 30 – 24 = 6.

9. Find the number of factors of the product 5⁸ * 7⁵ * 2³ which are perfect squares.
a) 47
b) 30
c) 65
d) 19

Answer: b
Clarification: Any factor of this number should be of the form 5^a * 7^b * 2^c. For the factor to be a perfect square a, b, c has to be even. a can take values 0, 2, 4, 6, 8, b can take values 0, 2, 4 and c can take values 0, 2. Total number of perfect squares = 5 * 3 * 2 = 30.

10. From a group of 8 men and 6 women, five persons are to be selected to form a committee so that at least 3 women are there on the committee. In how many ways can it be done?
a) 686
b) 438
c) 732
d) 549

Answer: a
Clarification: We may have (2 men and 3 women) or (1 men and 4 woman) or (5 women only). The Required number of ways = (⁸C₂ × ⁶C₃) + (⁸C₁ × ⁶C₄) + (⁶C₅) = 686.

Discrete Mathematics Multiple Choice Questions on “Discrete Probability – Logarithmic Series”.

1. Computation of the discrete logarithm is the basis of the cryptographic system _______
a) Symmetric cryptography
b) Asymmetric cryptography
c) Diffie-Hellman key exchange
d) Secret key cryptography

Answer: c
Clarification: A discrete logarithm modulo of an integer to the base is an integer such that a^x ≡ b (mod g). The problem of computing the discrete logarithm is a well-known challenge in the field of cryptography and is the basis of the cryptographic system i.e., the Diffie-Hellman key exchange.

2. Solve the logarithmic function of ln((frac{1+5x}{1+3x})).
a) 2x – 8x² + (frac{152x^3}{3}) – …
b) x² + (frac{7x^2}{2} – frac{12x^3}{5}) + …
c) x – (frac{15x^2}{2} + frac{163x^3}{4}) – …
d) 1 – (frac{x^2}{2} + frac{x^4}{4}) – …

Answer: a
Clarification: To solve the logarithmic function ln((frac{1+5x}{1+3x})) = ln(1+5x) – ln(1+3x) = (5x – (frac{(5x)^2}{2} + frac{(5x)^3}{3}) – …) – (3x – (frac{(3x)^2}{2} + frac{(3x)^3}{3}) – …) = 2x – 8x² + (frac{152x^3}{3}) – …

3. Determine the logarithmic function of ln(1+5x)^-5.
a) 5x + (frac{25x^2}{2} + frac{125x^3}{3} + frac{625x^4}{4}) …
b) x – (frac{25x^2}{2} + frac{625x^3}{3} – frac{3125x^4}{4}) …
c) (frac{125x^2}{3} – 625x^3 + frac{3125x^4}{5}) …
d) -25x + (frac{125x^2}{2} – frac{625x^3}{3} + frac{3125x^4}{4}) …

Answer: d
Clarification: Apply the logarithmic law, that is logax = xlog(a). Now the function is ln(1+5x)^-5 = -5log(1+5x). By taking the series = -5(5x – (frac{(5x)^2}{2} + frac{(5x)^3}{3} – frac{(5x)^4}{4}) + …) = -25x + (frac{125x^2}{2} – frac{625x^3}{3} + frac{3125x^4}{4}) …

4. Find the value of x: 3 x² a^log_ax = 348?
a) 7.1
b) 4.5
c) 6.2
d) 4.8

Answer: d
Clarification: Since, a^log_ax = x. The given equation may be written as: 3x² x = 348 ⇒ x = (116)^1/3 = 4.8.

5. Solve for x: log₂(x²-3x)=log₂(5x-15).
a) 2, 5
b) 7
c) 23
d) 3, 5

Answer: d
Clarification: By using the property if log_ax = log_ay then x=y, gives 2x²-3x=10-6x. Now, to solve the equation x²-3x-5x+15=0 ⇒ x²-8x+15 ⇒ x=3, x=5
For x=3: log₂(3²-3*3) = log₂(5*3-15) ⇒ true
For x=5: log₂(5²-3*5) = log₂(5*5-15) ⇒ true
The solutions to the equation are : x=3 and x=5.

6. Solve for x the equation 2^{x + 3} = 5^{x + 2}.
a) ln (24/8)
b) ln (25/8) / ln (2/5)
c) ln (32/5) / ln (2/3)
d) ln (3/25)

Answer: b
Clarification: Given that 2^{x + 3} = 5^{x + 2}. By taking ln of both sides: ln (2^{x + 3}) = ln (5^{x + 2})
⇒ (x + 3) ln 2 = (x + 2) ln 5
⇒ x ln 2 + 3 ln 2 = x ln 5 + 2 ln 5
⇒ x ln 2 – x ln 5 = 2 ln 5 – 3 ln 2
⇒ x = ( 2 ln 5 + 3 ln 2 ) / (ln 2 – ln 5) = ln (5² / 2³) / ln (2/5) = ln (25/8) / ln (2/5).

7. Given: log₄ z = B log_2/3z, for all z > 0. Find the value of constant B.
a) 2/(3!*ln(2))
b) 1/ln(7)
c) (4*ln(9))
d) 1/(2*ln(3))

Answer: d
Clarification: By using change of base formula we can have ln (x) / ln(4) = B ln(x) / ln(2/3) ⇒
B = 1/(2*ln(3)).

8. Evaluate: 16^x – 4^x – 9 = 0.
a) ln [( 5 + (sqrt{21})) / 2] / ln 8
b) ln [( 2 + (sqrt{33})) / 2] / ln 5
c) ln [( 1 + (sqrt{37})) / 2] / ln 4
d) ln [( 1 – (sqrt{37})) / 2] / ln 3

Answer: c
Clarification: Given: 16^x – 4^x – 9 = 0. Since 16^x = (4^x)², the equation may be written as: (4^x)² – 4^x – 9 = 0. Let t = 3^x and so t: t² – t – 9 = 0 which gives t: t = (1 + (sqrt{37})) / 2 and (1 – (sqrt{37})) / 2
Since t = 4x, the acceptable solution is y = (1 + (sqrt{37})) / 2 ⇒ 4x = (1 + (sqrt{37}))/2. By using ln on both sides: ln 4^x = ln [ (1 + (sqrt{37})) / 2] ⇒ x = ln [ ( 1 + (sqrt{37}))/2] / ln 3.

9. Transform 54^y = n+1 into equivalent a logarithmic expression.
a) log₁₂ (n+1)
b) log₄₁ (n²)
c) log₆₃ (n)
d) log₅₄ (n+1)

Answer: d
Clarification: By using the equivalent expression: a^y = x ⇔ y = log_a (x) to write 3^x = m as a logarithm: y = log₅₄ (n+1).

10. If log_a((frac{1}{8}) = -frac{3}{4}), than what is x?
a) 287
b) 469
c) 512
d) 623

Answer: c
Clarification: By using exponential form: a^-5/9 = 2/8. Now, raise both sides of the above equation to the power -9/5: (x^-5/9)^-9/5 = (1/32)^-9/5. By simplifying we get, a = 32^9/5 = 2⁹ = 512.

Discrete Mathematics Multiple Choice Questions on “Graph’s Matrices”.

1. A direct product of a group G possess which of the following characteristics?
a) a multiplication of subgroups of G
b) a factorization via subgroups of G
c) a superset of subgroups of G
d) a maximal power set of subgroups

Answer: b
Clarification: A direct product of a group G is a factorization via subgroups of G when the intersection is nontrivial, say X and Y, such that G = XY, X intersect Y = 1, and [X, Y]=1 and X, Y are normal in G.

2. In invariant algebra, some generators of group G1 that goes either into itself or zero under ______ with any other element of the algebra.
a) commutation
b) permutation
c) combination
d) lattice

Answer: a
Clarification: Some generators of group G1 in group theory which goes either into itself or zero under commutation with any other element of the whole algebra is called invariant subalgebra.

3. Which of the following can be embedded in an algebraically closed group?
a) infinite group
b) stargraph
c) a countable group
d) a semilattice

Answer: c
Clarification: We know that any countable group can always be embedded in an algebraically closed group.

4. Which of the following is the set of m×m invertible matrices?
a) a permutation group of degree m²
b) a general linear group of degree m
c) a sublattice group of degree m
d) a isomorphic graph of m nodes

Answer: b
Clarification: The general linear group of degree m is the set of m×m invertible matrices, consists of a general linear group of degree m having the ordinary matrix multiplication operation.

5. If any group is a manifold what is the dimension of that group?
a) same as manifold
b) same as vector space
c) infinite
d) finite

Answer: a
Clarification: If a group is a (topological) manifold, then the dimension of a group will be the dimension of this manifold. A linear representation F of a group G₁ on a vector space V’ has the dimension of V’.

6. A Latin square graph is a representation of a _______
a) quasi group
b) homomorphic group
c) semigroup
d) subgroup

Answer: a
Clarification: We know that any group is a representation of a graph. Now, a Quasi Group can be represented by a Latin Square matrix or by a Latin Square graph.

7. There exists _______ between group homology and group cohomology of a finite group.
a) homomorphism
b) isomorphism
c) automorphism
d) semilattice structure

Answer: a
Clarification: We know that there exists an isomorphism between group homology and group cohomology of finite group. Let S’ denote the set of all integers, and let G’ be a finite cyclic Group and for every S then G’-module N, we have S’S’_n(G’, A) is isomorphic to S’_n+1(G’, A).

8. In basic ring theory, any ring R1 may be embedded in its own ________
a) semilattice
b) endomorphism ring
c) homomorphic ring
d) subgroup

Answer: b
Clarification: We know that in basic ring theory, any ring R with its identity can be embedded in its own endomorphism ring and this is one of the most important characterization of rings. The endomorphism ring can contain a copy of its ring.

9. In Modern particle physics there must exist ______________
a) group theory
b) graph theory
c) lattice structure
d) invariant semigroup

Answer: a
Clarification: Modern particle physics exists with group theory. Group theory can predict the existence of many elementary particles. Depending on different symmetries, the structure and behaviour of molecules and crystals can be defined.

10. For any graph say G, Cayley graph is ______________
a) canonial
b) not canonical
c) isomorphic
d) homomorphic

Answer: b
Clarification: A different Cayley graph will be given for each choice of a generating set. Hence, the Cayley graph is not canonical.

Discrete Mathematics Puzzles on “Groups – Existence of Identity & Inverse”.

1. In a group there must be only __________ element.
a) 1
b) 2
c) 3
d) 5

Answer: a
Clarification: There can be only one identity element in a group and each element in a group has exactly one inverse element. Hence, two important consequences of the group axioms are the uniqueness of the identity element and the uniqueness of inverse elements.

2. _____ is the multiplicative identity of natural numbers.
a) 0
b) -1
c) 1
d) 2

Answer: c
Clarification: 1 is the multiplicative identity of natural numbers as a⋅1=a=1⋅a ∀a∈N. Thus, 1 is the identity of multiplication for the set of integers(Z), set of rational numbers(Q), and set of real numbers(R).

3. An identity element of a group has ______ element.
a) associative
b) commutative
c) inverse
d) homomorphic

Answer: c
Clarification: By the definition of all elements of a group have an inverse. For an element, a in a group G, an inverse of a is an element b such that ab=e, where e is the identity in the group. The inverse of an element is unique and usually denoted as -a.

4. __________ matrices do not have multiplicative inverses.
a) non-singular
b) singular
c) triangular
d) inverse

Answer: b
Clarification: The rational numbers are an extension of the integer numbers in which each non-zero number has an inverse under multiplication. A 3 × 3 matrix may or may not have an inverse under matrix multiplication. The matrices which do not have multiplicative inverses are termed as singular matrices.

5. If X is an idempotent nonsingular matrix, then X must be ___________
a) singular matrix
b) identity matrix
c) idempotent matrix
d) nonsingular matrix

Answer: b
Clarification: Since X is idempotent, we have X²=X. As X is nonsingular, it is invertible. Thus, the inverse matrix X^-1 exists. Then we have, I=X^-1X = X^-1X2=IX=X.

6. If A, B, and C are invertible matrices, the expression (AB^-1)^-1(CA^-1)^-1C2 evaluates to ____________
a) BC
b) C^-1BC
c) AB^-1
d) C^-1B

Answer: a
Clarification: Using the properties (AB)^-1=b^-1A^-1 and (A^-1)^-1=A, we may have,
(AB^-1)^-1(CA^-1)^-1C2
=(B^-1)^-1A^-1(A^-1)^-1C^-1C2
=BA^-1AC^-1C2
=BIC=BC [As, A^-1A=I].

7. If the sum of elements in each row of an n×n matrix Z is zero, then the matrix is ______________
a) inverse
b) non-singular
c) additive inverse
d) singular

Answer: d
Clarification: By the definition, an n×n matrix A is said to be singular if there exists a nonzero vector v such that Av=0. Otherwise, it is known that A is a nonsingular matrix.

8. ___________ are the symmetry groups used in the Standard model.
a) lie groups
b) subgroups
c) cyclic groups
d) poincare groups

Answer: a
Clarification: A symmetry group can encode symmetry features of a geometrical object. The group consists of the set of transformations that leave the object unchanged. Lie groups are such symmetry groups used in the standard model of particle physics.

9. A semigroup S under binary operation * that has an identity is called __________
a) multiplicative identity
b) monoid
c) subgroup
d) homomorphism

Answer: b
Clarification: Let P(S) is a commutative semigroup has the identity e, since e*A=A=A*e for any element A belongs to P(S). Hence, P(S) is a monoid.

10. An element a in a monoid is called an idempotent if ______________
a) a^-1=a*a^-1
b) a*a²=a
c) a²=a*a=a
d) a³=a*a

Answer: c
Clarification: An algebraic structure with a single associative binary operation and an Identity element are termed as a monoid. It is studied in semigroup theory. An element x in a monoid is called idempotent if a² = a*a = a.

Discrete Mathematics test on “Algebraic Laws on Sets”.

1. Let C and D be two sets then which of the following statements are true?

i) C U D = D U C                                          
ii) C ∩ D = D ∩ C

a) Both of the statements
b) Only i statement
c) Only ii statement
d) None of the statements

Answer: a
Clarification: Commutative laws hold good in sets.

2. If set C is {1, 2, 3, 4} and C – D = Φ then set D can be ___________
a) {1, 2, 4, 5}
b) {1, 2, 3}
c) {1, 2, 3, 4, 5}
d) None of the mentioned

Answer: c
Clarification: C ∩ D should be equivalent to C for C – D = Φ.

3. Let C and D be two sets then C – D is equivalent to __________
a) C’ ∩ D
b) C‘∩ D’
c) C ∩ D’
d) None of the mentioned

Answer: c
Clarification: Set C-D will be having those elements which are in C but not in D.

4. For two sets C and D the set (C – D) ∩ D will be __________
a) C
b) D
c) Φ
d) None of the mentioned

Answer: c
Clarification: C-D ≡ C ∩ D’, D ∩ D’ ≡ Φ.

5. Which of the following statement regarding sets is false?
a) A ∩ A = A
b) A U A = A
c) A – (B ∩ C) = (A – B) U (A –C)
d) (A U B)’ = A’ U B’

Answer: d
Clarification: (A U B)’ = A’ ∩ B’.

6. Let C = {1,2,3,4} and D = {1, 2, 3, 4} then which of the following hold not true in this case?
a) C – D = D – C
b) C U D = C ∩ D
c) C ∩ D = C – D
d) C – D = Φ

Answer: c
Clarification: C ∩ D = {1, 2, 3, 4} ≠ Φ.

7. If C’ U (D ∩ E’) is equivalent to __________
a) (C ∩ (D U E))’
b) (C ∩( D∩ E’))’
c) (C ∩( D’ U E))’
d) (C U ( D ∩ E’)’

Answer: c
Clarification: (C’)’≡ C, (C∩ D)’ ≡ C’ U D’.

8. Let Universal set U is {1, 2, 3, 4, 5, 6, 7, 8}, (Complement of A) A’ is {2, 5, 6, 7}, A ∩ B is {1, 3, 4} then the set B’ will surely have of which of the element?
a) 8
b) 7
c) 1
d) 3

Answer: a
Clarification: The set A is {1,3,4,8} and thus surely B does not have 8 in it. Since 8 does not belong to A ∩ B. For other element like 7 we can’t be sure.

9. Let a set be A then A ∩ φ and A U φ are __________
a) φ, φ
b) φ, A
c) A, φ
d)None of the mentioned

Answer: b
Clarification: By Domination Laws on sets.

10. If in sets A, B, C, the set B ∩ C consists of 8 elements, set A ∩ B consists of 7 elements and set C ∩ A consists of 7 elements then the minimum element in set A U B U C will be?
a) 8
b) 14
c) 22
d) 15

Answer: a
Clarification: For minimum elements set B and C have 8 elements each and all of the elements are same, Also set A should have 7 elements which are already present in B and C. Thus A U B U C ≡ A ≡ B.

Discrete Mathematics online test on “Operations on Matrices”.

1. Let A and B be two matrices of same order, then state whether the given statement is true or false.

A + B = B + A

a) True
b) False

Answer: a
Clarification: Matrix addition is commutative.

2. Let A and B be two matrices of same order, then state whether the given statement is true or false.

AB = BA

a) True
b) False

Answer: b
Clarification: Matrix multiplication is not commutative.

3. Let A order(axb) and Border(cxd) be two matrices, then for AB to exist, correct relation is given by?
a) a = d
b) b = c
c) a = b
d) c = d

Answer: b
Clarification: Matrix multiplication exists only when column of first matrix is same as rows of second i.e b = c.

4. Let A order(axb) and Border(cxd) be two matrices, then if AB exists, the order of AB is?
a) axd
b) bxc
c) axb
d) cxd

Answer: a
Clarification: Matrix multiplication exists only when column of first matrix is same as rows of second i.e b = c also resultant matrix will have number of rows equal to first matrix and column equal to the second matrix.

5. Let A=[a_ij ] be an mxn matrix and k be a scalar then kA is equal to __________
a) [ka_ij ]_mxn
b) [a_ij/k ]_mxn
c) [k² a_ij]_mxn
d) None of the mentioned

Answer: a
Clarification: The scalar is multiplied with each of the element of matrix A.

6. The matrix multiplication is distrbutive over matrix addition.
a) True
b) False

Answer: a
Clarification: For matrix A, B, C, A(B+C) = AB + AC.

7. If for a square matrix A, A² = A then such a matrix is known as _________
a) Idempotent matrix
b) Orthagonal matrix
c) Null matrix
d) None of the mentioned

Answer: a
Clarification: A sqaure matrix is called an Idempotent matrix, if A² = A.

8. For matrix A, B.(A+B)^T = A^T + B^T and (AB)^T = A^TB^T if the orders of matrices are appropriate.
a) True
b) False

Answer: b
Clarification: (A+B)^T = A^T + B^T is correct but (AB)^T = B^TA^T(reversal law).

9. For matrix A, B if A – B = O, where O is a null matrix then?
a) A = O
b) B = O
c) A = B
d) None of the mentioned

Answer: c
Clarification: If subtraction of B from A results in the null matrix this means that A is equivalent to B.

10. All the diagonal elements of a skew-symmetric matrix is?
a) 0
b) 1
c) 2
d) Any integer

Answer: a
Clarification: Since for a skew symmetric matrix a_ij = -a_ij, this implies all diagonal elements should be zero.