WGU C960 DISCRETE MATHEMATICS II: 200
PRACTICE EXAM QUESTIONS WITH
COMPLETE SOLUTIONS WITH VERIFIED
ANSWERS AND RATIONALES A+ GRADED
Counting & Combinatorics (Questions 1-28)
Q1. A password consists of 4 distinct letters from the English alphabet (a-z)
followed by 3 distinct digits (0-9). How many unique passwords can be created?
• A) 26⁴ × 10³
• B) P(26, 4) × P(10, 3)
• C) C(26, 4) × C(10, 3)
• D) 26 × 25 × 24 × 23 × 10 × 10 × 10
Answer: B
Rationale: Since both the letters and digits must be distinct and their order matters
(a password is an ordered sequence), we use permutations. The number of ways to
choose and arrange 4 distinct letters from 26 is P(26,4). The number of ways to
choose and arrange 3 distinct digits from 10 is P(10,3). By the product rule, the total
number of unique passwords is the product of these two values .
Q2. A committee of 5 members is to be formed from a group of 8 men and 6
women. If the committee must contain exactly 3 men, how many different
committees can be formed?
• A) C(8, 3) × C(6, 2)
• B) P(8, 3) × P(6, 2)
• C) C(14, 5)
• D) C(8, 3) + C(6, 2)
Answer: A
Rationale: The order in which committee members are chosen does not matter, so
we use combinations. First, choose 3 men from the 8 available: C(8,3) = 56. Since the
committee must have 5 members total, we need 2 women: C(6,2) = 15. Multiply to
get the total number of committees: 56 × 15 = 840 .
,Q3. How many 3-digit numbers can be formed from digits {1,2,3,4,5} with
repetition allowed?
• A) 60
• B) 125
• C) 120
• D) 15
Answer: B
Rationale: With repetition allowed, use the product rule: 5 choices for the hundreds
digit, 5 for the tens, and 5 for the ones. 5 × 5 × 5 = 125 .
Q4. How many 3-digit numbers can be formed from digits {1,2,3,4,5} without
repetition?
• A) 60
• B) 125
• C) 120
• D) 15
Answer: A
Rationale: Without repetition, use the product rule with decreasing choices: 5
choices for the hundreds digit, 4 for the tens (since one digit has been used), and 3
for the ones. 5 × 4 × 3 = 60. This is equivalent to P(5,3) .
Q5. What keywords indicate a combination problem should be used?
• A) "Line up," "arrange," "seat"
• B) "Choose," "select," "committee," "hand"
• C) "Password," "code," "string"
• D) "Either ... or ..."
Answer: B
Rationale: When a problem asks you to choose a group where order does not
matter, use combinations. Keywords include "choose," "select," "committee," and
"hand." Use permutations when seeing "line up," "arrange," "seat," or "rank" .
Q6. When should the product rule be used in counting problems?
• A) When the problem says "either ... or ..."
• B) When filling positions left to right with independent choices
• C) When order does not matter
• D) When there is symmetry to correct for
,Answer: B
Rationale: The product rule states that if one event can occur in m ways and a
second independent event can occur in n ways, the total number of ways the two
events can occur in sequence is m × n. Use this when creating passwords, codes, or
arranging items with independent position choices .
Q7. When should the sum rule be used in counting problems?
• A) When the problem says "either ... or ..."
• B) When filling positions left to right
• C) When order does not matter
• D) When there is symmetry to correct for
Answer: A
Rationale: The sum rule states that if two tasks cannot be done simultaneously and
the first can be done in m ways and the second in n ways, there are m + n total ways
to perform one of these tasks. Solve each case and add the results .
Q8. What is the cardinality of the power set of a set with 5 elements?
• A) 25
• B) 32
• C) 10
• D) 5
Answer: B
Rationale: The cardinality of the power set is 2^n, where n is the number of
elements in the original set. Therefore, 2^5 = 32 .
Q9. What does the division rule correct for in counting?
• A) Missing a case
• B) Overcounting due to symmetry, rotations, or identical items
• C) Using the wrong formula
• D) Forgetting to multiply
Answer: B
Rationale: The division rule corrects for k-to-1 overcounting. Divide your count by k,
where your count is the number obtained by counting normally and k is how many
times each distinct outcome was counted .
Q10. How many strings of 8 bits either start with 1 or end with the two bits 00?
, • A) 128
• B) 160
• C) 192
• D) 224
Answer: B
Rationale: Use inclusion-exclusion:
• Strings starting with 1: 2^7 = 128
• Strings ending with 00: 2^6 = 64
• Strings doing both: 2^5 = 32
Total = 128 + 64 - 32 = 160 .
Q11. In how many ways can 7 distinct books be arranged on a shelf?
• A) 5040
• B) 720
• C) 49
• D) 7
Answer: A
Rationale: Arranging 7 distinct items in order is a permutation of 7 items: 7! = 5040.
Q12. In how many ways can 4 people be seated in a row of 6 chairs?
• A) 360
• B) 15
• C) 24
• D) 720
Answer: A
Rationale: This is a permutation of 6 items taken 4 at a time: P(6,4) = 6 × 5 × 4 × 3 =
360.
Q13. How many ways are there to choose 2 vowels and 3 consonants from the
English alphabet?
• A) C(5,2) × C(21,3)
• B) P(5,2) × P(21,3)
• C) C(26,5)
• D) 5 × 21
Answer: A
PRACTICE EXAM QUESTIONS WITH
COMPLETE SOLUTIONS WITH VERIFIED
ANSWERS AND RATIONALES A+ GRADED
Counting & Combinatorics (Questions 1-28)
Q1. A password consists of 4 distinct letters from the English alphabet (a-z)
followed by 3 distinct digits (0-9). How many unique passwords can be created?
• A) 26⁴ × 10³
• B) P(26, 4) × P(10, 3)
• C) C(26, 4) × C(10, 3)
• D) 26 × 25 × 24 × 23 × 10 × 10 × 10
Answer: B
Rationale: Since both the letters and digits must be distinct and their order matters
(a password is an ordered sequence), we use permutations. The number of ways to
choose and arrange 4 distinct letters from 26 is P(26,4). The number of ways to
choose and arrange 3 distinct digits from 10 is P(10,3). By the product rule, the total
number of unique passwords is the product of these two values .
Q2. A committee of 5 members is to be formed from a group of 8 men and 6
women. If the committee must contain exactly 3 men, how many different
committees can be formed?
• A) C(8, 3) × C(6, 2)
• B) P(8, 3) × P(6, 2)
• C) C(14, 5)
• D) C(8, 3) + C(6, 2)
Answer: A
Rationale: The order in which committee members are chosen does not matter, so
we use combinations. First, choose 3 men from the 8 available: C(8,3) = 56. Since the
committee must have 5 members total, we need 2 women: C(6,2) = 15. Multiply to
get the total number of committees: 56 × 15 = 840 .
,Q3. How many 3-digit numbers can be formed from digits {1,2,3,4,5} with
repetition allowed?
• A) 60
• B) 125
• C) 120
• D) 15
Answer: B
Rationale: With repetition allowed, use the product rule: 5 choices for the hundreds
digit, 5 for the tens, and 5 for the ones. 5 × 5 × 5 = 125 .
Q4. How many 3-digit numbers can be formed from digits {1,2,3,4,5} without
repetition?
• A) 60
• B) 125
• C) 120
• D) 15
Answer: A
Rationale: Without repetition, use the product rule with decreasing choices: 5
choices for the hundreds digit, 4 for the tens (since one digit has been used), and 3
for the ones. 5 × 4 × 3 = 60. This is equivalent to P(5,3) .
Q5. What keywords indicate a combination problem should be used?
• A) "Line up," "arrange," "seat"
• B) "Choose," "select," "committee," "hand"
• C) "Password," "code," "string"
• D) "Either ... or ..."
Answer: B
Rationale: When a problem asks you to choose a group where order does not
matter, use combinations. Keywords include "choose," "select," "committee," and
"hand." Use permutations when seeing "line up," "arrange," "seat," or "rank" .
Q6. When should the product rule be used in counting problems?
• A) When the problem says "either ... or ..."
• B) When filling positions left to right with independent choices
• C) When order does not matter
• D) When there is symmetry to correct for
,Answer: B
Rationale: The product rule states that if one event can occur in m ways and a
second independent event can occur in n ways, the total number of ways the two
events can occur in sequence is m × n. Use this when creating passwords, codes, or
arranging items with independent position choices .
Q7. When should the sum rule be used in counting problems?
• A) When the problem says "either ... or ..."
• B) When filling positions left to right
• C) When order does not matter
• D) When there is symmetry to correct for
Answer: A
Rationale: The sum rule states that if two tasks cannot be done simultaneously and
the first can be done in m ways and the second in n ways, there are m + n total ways
to perform one of these tasks. Solve each case and add the results .
Q8. What is the cardinality of the power set of a set with 5 elements?
• A) 25
• B) 32
• C) 10
• D) 5
Answer: B
Rationale: The cardinality of the power set is 2^n, where n is the number of
elements in the original set. Therefore, 2^5 = 32 .
Q9. What does the division rule correct for in counting?
• A) Missing a case
• B) Overcounting due to symmetry, rotations, or identical items
• C) Using the wrong formula
• D) Forgetting to multiply
Answer: B
Rationale: The division rule corrects for k-to-1 overcounting. Divide your count by k,
where your count is the number obtained by counting normally and k is how many
times each distinct outcome was counted .
Q10. How many strings of 8 bits either start with 1 or end with the two bits 00?
, • A) 128
• B) 160
• C) 192
• D) 224
Answer: B
Rationale: Use inclusion-exclusion:
• Strings starting with 1: 2^7 = 128
• Strings ending with 00: 2^6 = 64
• Strings doing both: 2^5 = 32
Total = 128 + 64 - 32 = 160 .
Q11. In how many ways can 7 distinct books be arranged on a shelf?
• A) 5040
• B) 720
• C) 49
• D) 7
Answer: A
Rationale: Arranging 7 distinct items in order is a permutation of 7 items: 7! = 5040.
Q12. In how many ways can 4 people be seated in a row of 6 chairs?
• A) 360
• B) 15
• C) 24
• D) 720
Answer: A
Rationale: This is a permutation of 6 items taken 4 at a time: P(6,4) = 6 × 5 × 4 × 3 =
360.
Q13. How many ways are there to choose 2 vowels and 3 consonants from the
English alphabet?
• A) C(5,2) × C(21,3)
• B) P(5,2) × P(21,3)
• C) C(26,5)
• D) 5 × 21
Answer: A
