Finite Mathematics Quiz: Addition And Multiplication Principles

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Addition And Multiplication PrinciplesQuestion 1 of 20

A student must take exactly 7 courses to graduate: 3 mathematics courses chosen from 6 available, 2 science courses chosen from 5 available, and 2 electives chosen from 8 available. However, if the student takes Advanced Calculus (one of the 6 math courses), then they cannot take Physics II (one of the 5 science courses) due to scheduling conflicts. How many different course combinations are possible?

\binom{6}{3} \times \binom{5}{2} \times \binom{8}{2} - \binom{5}{2} \times \binom{4}{1} \times \binom{8}{2}

\binom{6}{3} \times \binom{5}{2} \times \binom{8}{2} - \binom{5}{2} \times \binom{4}{2} \times \binom{8}{2}

\binom{5}{3} \times \binom{5}{2} \times \binom{8}{2} + \binom{5}{2} \times \binom{4}{2} \times \binom{8}{2}

\binom{6}{3} \times \binom{5}{2} \times \binom{8}{2} - \binom{5}{2} \times \binom{4}{2} \times \binom{7}{2}

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Finite Mathematics Quiz

Finite Mathematics Quiz: Addition And Multiplication Principles

Practice Addition And Multiplication Principles in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Addition And Multiplication Principles, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$\binom{6}{3} \times \binom{5}{2} \times \binom{8}{2} - \binom{5}{2} \times \binom{4}{1} \times \binom{8}{2}$ (correct answer)
$\binom{6}{3} \times \binom{5}{2} \times \binom{8}{2} - \binom{5}{2} \times \binom{4}{2} \times \binom{8}{2}$
$\binom{5}{3} \times \binom{5}{2} \times \binom{8}{2} + \binom{5}{2} \times \binom{4}{2} \times \binom{8}{2}$
$\binom{6}{3} \times \binom{5}{2} \times \binom{8}{2} - \binom{5}{2} \times \binom{4}{2} \times \binom{7}{2}$

Explanation: Use inclusion-exclusion principle. Total combinations without restrictions:

\binom{6}{3} \times \binom{5}{2} \times \binom{8}{2}

. Subtract combinations where both Advanced Calculus and Physics II are taken. If Advanced Calculus is taken, we need 2 more math courses from the remaining 5, and if Physics II is taken, we need 1 more science course from the remaining 4. This gives us the subtraction term:

\binom{5}{2} \times \binom{4}{1} \times \binom{8}{2}

. Choice B incorrectly uses

\binom{4}{2}

for remaining science courses. Choice C uses addition instead of subtraction and has incorrect math course counting. Choice D incorrectly affects elective choices.

Question 2

A restaurant offers a prix fixe menu where customers must choose exactly one item from each of four categories: appetizer (6 options), soup (4 options), main course (8 options), and dessert (5 options). However, certain combinations are not available: if a customer chooses the lobster appetizer, then the fish main course is not available, and if a customer chooses the chocolate dessert, then the wine soup is not available. Assuming these are the only restrictions, how many different meal combinations are possible?

$6 \times 4 \times 8 \times 5 - 1 \times 1 \times 4 \times 5 - 6 \times 1 \times 8 \times 1$
$6 \times 4 \times 8 \times 5 - 1 \times 4 \times 1 \times 5 - 6 \times 1 \times 8 \times 1 + 1 \times 1 \times 1 \times 1$ (correct answer)
$6 \times 4 \times 8 \times 5 - 1 \times 4 \times 1 \times 5 - 6 \times 1 \times 8 \times 1$
$5 \times 4 \times 7 \times 5 + 1 \times 3 \times 8 \times 5 + 6 \times 4 \times 8 \times 4$

Explanation: Use inclusion-exclusion principle. Total combinations without restrictions:

6 \times 4 \times 8 \times 5 = 960

. Subtract forbidden combinations: (lobster appetizer AND fish main course):

1 \times 4 \times 1 \times 5 = 20

. Also subtract: (wine soup AND chocolate dessert):

6 \times 1 \times 8 \times 1 = 48

. But we've double-subtracted the case where all three restricted items are chosen together (lobster appetizer AND fish main course AND wine soup AND chocolate dessert), so we must add it back:

1 \times 1 \times 1 \times 1 = 1

. Final answer:

960 - 20 - 48 + 1 = 893

. Choice A omits the inclusion step. Choice C incorrectly applies inclusion-exclusion. Choice D uses a completely different approach.

Question 3

A license plate consists of 3 letters followed by 4 digits. How many license plates can be formed if no letter or digit can be repeated within the same license plate, and the license plate cannot begin with the letters I, O, or Q?

$23 \times 25 \times 24 \times 10 \times 9 \times 8 \times 7$ (correct answer)
$23 \times 26 \times 25 \times 10 \times 9 \times 8 \times 7$
$23 \times 25 \times 24 \times 9 \times 8 \times 7 \times 6$
$26 \times 25 \times 24 \times 10 \times 9 \times 8 \times 7$

Explanation: The first letter cannot be I, O, or Q, so there are 26 - 3 = 23 choices. The second letter can be any letter except the one already used (including I, O, or Q if not used first), giving 25 choices. The third letter has 24 remaining choices. For digits, we have 10 choices for the first digit, 9 for the second, 8 for the third, and 7 for the fourth. Choice B incorrectly gives 26 choices for the second letter. Choice C incorrectly starts with 9 choices for the first digit. Choice D ignores the restriction on the first letter.

Question 4

A committee of three distinct positions (President, VP, Treasurer) is to be formed from a group of 4 managers and 6 analysts. For the committee to be valid, the President must be a manager or the Treasurer must be an analyst, but not both. How many different valid committees can be formed?

192
336 (correct answer)
528
720

Explanation: This problem requires applying the addition principle to two mutually exclusive cases. The condition is an exclusive OR. Case 1: The President is a manager AND the Treasurer is NOT an analyst (meaning the Treasurer is also a manager). There are 4 choices for President (managers). Then there are 3 choices for Treasurer (remaining managers). There are 8 people left for the VP position. Ways for Case 1:

4 \times 3 \times 8 = 96

. Case 2: The President is NOT a manager (is an analyst) AND the Treasurer IS an analyst. There are 6 choices for President (analysts). Then there are 5 choices for Treasurer (remaining analysts). There are 8 people left for the VP position. Ways for Case 2:

6 \times 5 \times 8 = 240

. The total number of valid committees is the sum of the ways for these two cases:

96 + 240 = 336

Question 5

A standard license plate consists of 3 uppercase letters followed by 3 digits. If repetition of letters and digits is allowed, how many license plates contain at least one letter 'X' or at least one digit '7'?

6,713,096
11,390,625
6,185,375 (correct answer)
17,576,000

Explanation: This problem is best solved using the complement rule. First, find the total number of possible license plates without any restrictions. There are 26 choices for each of the 3 letters and 10 choices for each of the 3 digits. Total plates =

26^3 \times 10^3 = 17,576,000

. The complement of 'at least one X or at least one 7' is 'no X's AND no 7's'. The number of plates with no 'X's and no '7's is calculated using 25 letters and 9 digits. Number of complement plates =

25^3 \times 9^3 = 15,625 \times 729 = 11,390,625

. The number of plates that satisfy the condition is Total - Complement =

17,576,000 - 11,390,625 = 6,185,375

Question 6

A bookshelf holds 5 distinct math books, 4 distinct physics books, and 3 distinct chemistry books. How many ways can the books be arranged on the shelf if all the books of the same subject must be kept together?

17,280
156
103,680 (correct answer)
479,001,600

Explanation: This problem involves two levels of arrangement. First, treat each subject's books as a single block. There are 3 blocks (Math, Physics, Chemistry). These 3 blocks can be arranged in

3! = 6

ways. Second, arrange the books within each block. The 5 math books can be arranged in

5! = 120

ways. The 4 physics books can be arranged in

4! = 24

ways. The 3 chemistry books can be arranged in

3! = 6

ways. By the multiplication principle, the total number of arrangements is the product of these values:

3! \times 5! \times 4! \times 3! = 6 \times 120 \times 24 \times 6 = 103,680

Question 7

A club with 6 seniors and 4 juniors must form a 3-person executive committee with distinct roles: President, VP, and Secretary. The committee must contain at least one senior and at least one junior. How many such committees can be formed?

144
576 (correct answer)
600
720

Explanation: This problem is best solved using the complement rule. First, find the total number of ways to form the committee from the 10 members without any restrictions. This is a permutation:

P(10, 3) = 10 \times 9 \times 8 = 720

. The complement of 'at least one senior and at least one junior' is 'all seniors' or 'all juniors'. Case 1 (bad): All seniors. The committee is chosen from 6 seniors:

P(6, 3) = 6 \times 5 \times 4 = 120

. Case 2 (bad): All juniors. The committee is chosen from 4 juniors:

P(4, 3) = 4 \times 3 \times 2 = 24

. Using the addition principle, the total number of 'bad' committees is

120 + 24 = 144

. The number of 'good' committees is Total - Bad =

720 - 144 = 576

Question 8

A manager is creating an on-call schedule for 5 employees (A, B, C, D, E) for a 5-day work week (Monday-Friday). Each employee is assigned to exactly one day. Employee A must be on-call on either Monday or Tuesday. Employee B must be on-call on Friday. How many different schedules are possible?

6
12 (correct answer)
24
36

Explanation: This is a multiplication principle problem where we should fill the most constrained positions first. Friday's position has only 1 choice: Employee B. Employee A has 2 choices for their day: Monday or Tuesday. Once A's day is chosen and B's day is fixed, there are 3 employees (C, D, E) and 3 remaining days. The number of ways to assign these 3 employees to the 3 remaining days is a permutation:

3! = 3 \times 2 \times 1 = 6

. The total number of possible schedules is the product of the number of choices for each step: (Choices for A's day) × (Choices for B's day) × (Ways to arrange the rest) =

2 \times 1 \times 6 = 12

Question 9

A restaurant offers a dinner special where a customer chooses one appetizer, one entrée, and one dessert. There are 6 appetizers, 8 entrées, and 4 desserts available. Two of the appetizers and three of the entrées are vegetarian. A customer wants to order a meal where either the appetizer is vegetarian or the entrée is vegetarian, but not both. How many meal combinations are possible?

40
88 (correct answer)
112
192

Explanation: This problem uses the addition principle for two disjoint cases, with the multiplication principle applied within each. The condition 'but not both' indicates an exclusive OR. We have: 2 vegetarian appetizers,

6-2=4

non-vegetarian appetizers. 3 vegetarian entrées,

8-3=5

non-vegetarian entrées. 4 desserts. Case 1: Vegetarian appetizer AND non-vegetarian entrée. Number of choices:

2 \text{ (appetizer)} \times 5 \text{ (entrée)} \times 4 \text{ (dessert)} = 40

meals. Case 2: Non-vegetarian appetizer AND vegetarian entrée. Number of choices:

4 \text{ (appetizer)} \times 3 \text{ (entrée)} \times 4 \text{ (dessert)} = 48

meals. The total number of qualifying meals is the sum of these two mutually exclusive cases:

40 + 48 = 88

Question 10

A committee of 6 people is to be formed from a group of 8 men and 7 women. The committee must have at least 2 men and at least 2 women, and exactly one of the members must serve as chairperson. If the chairperson must be a woman, how many different committees are possible?

$7 \times \left[\binom{8}{3}\binom{6}{2} + \binom{8}{2}\binom{6}{3} + \binom{8}{1}\binom{6}{4}\right]$
$\binom{8}{2}\binom{7}{4} + \binom{8}{3}\binom{7}{3} + \binom{8}{4}\binom{7}{2}$
$7 \times \left[\binom{8}{2}\binom{6}{3} + \binom{8}{3}\binom{6}{2} + \binom{8}{4}\binom{6}{1}\right]$ (correct answer)
$\binom{15}{6} - \binom{8}{5}\binom{7}{1} - \binom{8}{1}\binom{7}{5} - \binom{8}{6}\binom{7}{0}$

Explanation: Since the chairperson must be a woman, we first choose 1 woman from 7 to be chairperson. Then we need to choose 5 more people from the remaining 14 people (8 men + 6 women), with constraints: at least 2 men and at least 1 more woman (since we need at least 2 women total). The valid combinations for the remaining 5 spots are: (2 men, 3 women), (3 men, 2 women), or (4 men, 1 woman). This gives us

7 \times \left[\binom{8}{2}\binom{6}{3} + \binom{8}{3}\binom{6}{2} + \binom{8}{4}\binom{6}{1}\right]

. Choice A has wrong gender counts, choice B forgets to account for chairperson selection, choice D uses inclusion-exclusion incorrectly.

Question 11

A student is answering a 5-question multiple-choice quiz where each question has 4 options. The student knows that no two consecutive questions have the same answer letter. How many different ways can the student answer the quiz?

243
1024
16
324 (correct answer)

Explanation: This problem is a direct application of the multiplication principle with a constraint. For the first question, there are 4 possible answers. For the second question, the answer cannot be the same as the first, leaving 3 possible answers. For the third question, the answer cannot be the same as the second, again leaving 3 possible answers. This pattern continues for the remaining questions. The total number of ways to answer the quiz is the product of the number of choices for each question:

4 \times 3 \times 3 \times 3 \times 3 = 4 \times 3^4 = 4 \times 81 = 324

Question 12

A website requires a 6-character password. The first three characters must be distinct uppercase letters from the English alphabet, and the last three characters must be distinct digits from 0 to 9. The first letter cannot be 'A' or 'B', and the last digit must be even. How many different passwords can be created?

5,184,000 (correct answer)
5,616,000
7,200,000
10,368,000

Explanation: This is a multi-step problem using the multiplication principle. First, calculate the number of ways to choose the letters. The first letter has 24 choices (26 total, excluding 'A' and 'B'). The second letter has 25 choices (any letter except the first). The third letter has 24 choices (any letter except the first two). So there are

24 \times 25 \times 24 = 14,400

ways to choose the letters. Next, calculate the number of ways to choose the digits. The last digit must be even, giving 5 choices ({0, 2, 4, 6, 8}). The first digit (fourth character overall) can be any of the 9 remaining digits. The second digit (fifth character) can be any of the 8 remaining digits. So there are

5 \times 9 \times 8 = 360

ways to choose the digits. The total number of passwords is the product of these two results:

14,400 \times 360 = 5,184,000

Question 13

A code consists of 4 symbols in order: two letters followed by two digits. The letters must be distinct and chosen from {A, B, C, D}. The digits must be distinct and chosen from {1, 2, 3}. A code is considered valid if the first letter is a vowel OR the last digit is even. How many different valid codes can be formed?

6
36 (correct answer)
42
72

Explanation: This problem requires the Principle of Inclusion-Exclusion:

|A \cup B| = |A| + |B| - |A \cap B|

. Let A be the set of codes where the first letter is a vowel, and B be the set of codes where the last digit is even. The only vowel is 'A', and the only even digit is '2'. Calculate

|A|

: If the first letter is 'A' (1 choice), the second letter has 3 choices from {B, C, D}. The two distinct digits can be arranged in

P(3, 2) = 3 \times 2 = 6

ways. So,

|A| = 1 \times 3 \times 6 = 18

. Calculate

|B|

: If the last digit is '2' (1 choice), the second-to-last digit has 2 choices from {1, 3}. The two distinct letters can be arranged in

P(4, 2) = 4 \times 3 = 12

ways. So,

|B| = 12 \times 2 \times 1 = 24

. Calculate

|A \cap B|

(the overlap): The first letter is 'A' AND the last digit is '2'. The second letter has 3 choices. The second-to-last digit has 2 choices. So,

|A \cap B| = 1 \times 3 \times 2 \times 1 = 6

. Total valid codes =

|A| + |B| - |A \cap B| = 18 + 24 - 6 = 36

Question 14

A board of directors is selecting a President, a Vice President, and a Treasurer from a group of 5 executives and 10 managers. How many ways can these three distinct positions be filled if at least one executive must be selected?

720
2,010 (correct answer)
2,160
2,730

Explanation: This problem is best solved using the principle of complementary counting. We find the total number of ways to fill the positions without any restrictions and then subtract the number of ways the condition is not met (the complement). The total number of people is

5 + 10 = 15

. Step 1: Calculate the total number of ways to select the three officers without any restrictions.

President: 15 choices.
Vice President: 14 choices (one person is already President).
Treasurer: 13 choices (two people are already selected). Total arrangements = $15 \times 14 \times 13 = 2,730$ .

Step 2: Calculate the number of ways to fill the positions that violate the condition. The condition is 'at least one executive is selected'. The violation is 'zero executives are selected', which means all three officers are managers.

President (must be a manager): 10 choices.
Vice President (must be a manager): 9 choices.
Treasurer (must be a manager): 8 choices. Arrangements with no executives = $10 \times 9 \times 8 = 720$ .

Step 3: Subtract the complement from the total. The number of ways with at least one executive = (Total arrangements) - (Arrangements with no executives).

2,730 - 720 = 2,010

. Distractor Rationale:

A is the number of arrangements with no executives (the complement).
C is the result of an incorrect attempt at direct counting, such as $5 \times 10 \times 9 + 5 \times 4 \times 10 + 5 \times 4 \times 3 = 450 + 200 + 60 = 710$ , which is not a choice, or a more common error: $3 \times 5 \times 10 \times 9 = 1350$ (1 exec, 2 managers) + $3 \times 5 \times 4 \times 10 = 600$ (2 execs, 1 manager) + $5 \times 4 \times 3 = 60$ (3 execs) = 2010. Let me check C again. $2160$ could be $6 \times 10 \times 9 imes 4$ No. Maybe $P(10,3) imes 3 = 720 imes 3=2160$ . This could be reasoning that there are 3 positions an executive could take. This is flawed reasoning.
D is the total number of arrangements without any restrictions.

Question 15

How many integers from 1 to 600, inclusive, are divisible by 4 or by 6?

150
200 (correct answer)
250
300

Explanation: This problem uses the Principle of Inclusion-Exclusion. Let

A

be the set of integers from 1 to 600 divisible by 4, and

B

be the set of integers divisible by 6. We want to find the size of the union,

|A \cup B|

, which is given by

|A| + |B| - |A \cap B|

Find $|A|$ , the number of integers divisible by 4: $|A| = \lfloor \frac{600}{4} \rfloor = 150$ .
Find $|B|$ , the number of integers divisible by 6: $|B| = \lfloor \frac{600}{6} \rfloor = 100$ .
Find $|A \cap B|$ , the number of integers divisible by both 4 and 6. An integer is divisible by both if it is divisible by their least common multiple, LCM(4, 6) = 12. $|A \cap B| = \lfloor \frac{600}{12} \rfloor = 50$ .
Apply the principle: $|A \cup B| = |A| + |B| - |A \cap B| = 150 + 100 - 50 = 200$ .

Distractor Rationale:

A is the count of numbers divisible by 4 only, $|A|$ , without considering those divisible by 6.
C is the result of adding the counts for 4 and 6 without subtracting the overlap ( $150 + 100 = 250$ ), a very common error.
D is the result of incorrectly adding the overlap instead of subtracting it ( $150 + 100 + 50 = 300$ ).

Question 16

A student is scheduling their four distinct final exams—Calculus, History, Physics, and Literature—on four consecutive days from Monday to Thursday. If the Calculus exam must be scheduled at some point before the Physics exam, how many different valid schedules are possible?

6
12 (correct answer)
18
24

Explanation: This problem can be solved with a symmetry argument or by casework. Method 1: Symmetry First, calculate the total number of ways to schedule the four distinct exams without any restrictions. This is a simple permutation.

Monday: 4 choices
Tuesday: 3 choices
Wednesday: 2 choices
Thursday: 1 choice Total schedules = $4 \times 3 \times 2 \times 1 = 4! = 24$ .

In any given schedule, the Calculus exam is either before the Physics exam or after it. These two possibilities are mutually exclusive and, by symmetry, equally likely. Therefore, exactly half of the total schedules will have Calculus before Physics. Number of valid schedules =

\frac{24}{2} = 12

. Method 2: Casework (Addition Principle) We consider the possible days for the Calculus exam.

Case 1: Calculus is on Monday (1 choice). Physics can be on Tue, Wed, or Thu (3 choices). The other two exams (History, Literature) can be arranged in the remaining 2 slots in $2! = 2$ ways. Total: $1 \times 3 \times 2 = 6$ ways.
Case 2: Calculus is on Tuesday (1 choice). Physics can be on Wed or Thu (2 choices). The other two exams can be arranged in the remaining 2 slots (Mon, the other day) in $2! = 2$ ways. Total: $1 \times 2 \times 2 = 4$ ways.
Case 3: Calculus is on Wednesday (1 choice). Physics must be on Thursday (1 choice). The other two exams can be arranged in the remaining 2 slots (Mon, Tue) in $2! = 2$ ways. Total: $1 \times 1 \times 2 = 2$ ways. (Calculus cannot be on Thursday, as it must be before Physics). Total valid schedules = $6 + 4 + 2 = 12$ .

Distractor Rationale:

A is the number of possibilities if Calculus is on Monday, only one of the cases.
D is the total number of unrestricted schedules.
C is a plausible guess but does not correspond to a standard calculation error.

Question 17

A valid password must be 5 characters long, using only the uppercase letters {A, B, C} and the digits {1, 2, 3, 4}. How many such passwords contain at least one letter and at least one digit?

1,267
15,540 (correct answer)
15,783
16,807

Explanation: This problem is best solved using complementary counting. First, we find the total number of possible passwords, then we subtract the number of 'invalid' passwords (those that do not meet the condition). The set of available characters has

3

letters and

4

digits, for a total of

3+4=7

characters. Repetition is allowed. Step 1: Calculate the total number of 5-character passwords without any restrictions. Since there are 7 choices for each of the 5 positions, the total is

7^5 = 16,807

. Step 2: Identify and count the passwords that do NOT meet the condition. The condition is 'at least one letter and at least one digit'. The complement is passwords that have 'NO letters' (i.e., all digits) OR 'NO digits' (i.e., all letters).

Passwords with NO letters (all digits): There are 4 available digits. The number of such passwords is $4^5 = 1,024$ .
Passwords with NO digits (all letters): There are 3 available letters. The number of such passwords is $3^5 = 243$ . These two sets are disjoint (a password cannot be both all-digits and all-letters).

Step 3: Calculate the total number of invalid passwords by adding the counts from Step 2. Total invalid passwords =

1,024 + 243 = 1,267

. Step 4: Subtract the invalid count from the total count. Valid passwords = Total passwords - Invalid passwords =

16,807 - 1,267 = 15,540

. Distractor Rationale:

A is the number of invalid passwords (the complement), not the number of valid ones.
C is the result of subtracting only one of the invalid cases from the total ( $16,807 - 243 = 16,564$ or $16,807 - 1,024 = 15,783$ ).
D is the total number of possible passwords without any restrictions.

Question 18

A restaurant offers a dinner special. For an appetizer, a customer can choose from soup (3 kinds) or salad (2 kinds). For the main course, if soup was chosen, there are 4 main course options. If salad was chosen, there are 6 main course options. Finally, every meal comes with one of 3 desserts. How many different dinner special combinations are possible?

24
72 (correct answer)
78
150

Explanation: We must consider two separate cases based on the appetizer choice. We use the Multiplication Principle within each case and the Addition Principle to combine the cases. Case 1: The customer chooses soup.

Appetizer choices (soup): 3
Main course choices (given soup): 4
Dessert choices: 3 Number of meal combinations in this case is $3 \times 4 \times 3 = 36$ .

Case 2: The customer chooses salad.

Appetizer choices (salad): 2
Main course choices (given salad): 6
Dessert choices: 3 Number of meal combinations in this case is $2 \times 6 \times 3 = 36$ .

Since these two cases are mutually exclusive (the customer cannot choose both soup and salad), the total number of different dinner combinations is the sum of the combinations from each case: Total combinations =

36 + 36 = 72

. Distractor Rationale:

A is the result of forgetting the dessert option ( $3 \times 4 + 2 \times 6 = 12 + 12 = 24$ ).
C could come from incorrectly associating the number of main courses with the wrong appetizer (3 soup kinds with 6 mains, 2 salad kinds with 4 mains), leading to $(3 \times 6 \times 3) + (2 \times 4 \times 3) = 54 + 24 = 78$ .
D results from incorrectly combining the choices for appetizers and main courses before multiplying: $(3+2) \times (4+6) \times 3 = 5 \times 10 \times 3 = 150$ . This reflects a misunderstanding of how the choices are dependent.

Question 19

A website's security protocol allows for two types of passwords: either a sequence of four distinct lowercase letters or a sequence of two distinct lowercase letters followed by two distinct digits (0-9). How many different passwords are possible?

58,500
358,800
417,300 (correct answer)
524,576

Explanation: This problem involves two mutually exclusive cases, so we calculate the number of possibilities for each case using the Multiplication Principle and then add the results using the Addition Principle. Case 1: The password is four distinct lowercase letters.

The number of choices for the first letter is 26.
The number of choices for the second letter is 25 (since it must be different from the first).
The number of choices for the third letter is 24.
The number of choices for the fourth letter is 23. Total for Case 1: $26 \times 25 \times 24 \times 23 = 358,800$ .

Case 2: The password is two distinct lowercase letters followed by two distinct digits.

The number of choices for the first letter is 26.
The number of choices for the second letter is 25.
The number of choices for the first digit is 10.
The number of choices for the second digit is 9. Total for Case 2: $26 \times 25 \times 10 \times 9 = 58,500$ .

Since the password can be from either Case 1 OR Case 2, we add the totals: Total possible passwords =

358,800 + 58,500 = 417,300

. Distractor Rationale:

A is the result for Case 2 only.
B is the result for Case 1 only.
D is the result if the 'distinct' constraint is ignored ( $26^4 + 26^2 \times 10^2 = 456,976 + 67,600 = 524,576$ ).

Question 20

A student needs to select a single research project. They can choose one from a list of 8 approved chemistry topics or one from a list of 5 approved physics topics. As a third option, the student can propose a special interdisciplinary topic. To do this, they must get approval from one of 3 specific professors, each of whom will approve one of 2 pre-defined project frameworks. How many total project options does the student have?

18
19 (correct answer)
46
240

Explanation: This problem requires identifying distinct categories of choices and applying the Addition and Multiplication Principles correctly. The student is making a single choice from three mutually exclusive sets of options: chemistry projects, physics projects, or special projects.

Number of Chemistry project options: 8.
Number of Physics project options: 5.
Number of special interdisciplinary project options: To determine this, we use the Multiplication Principle. The student must choose a professor AND a framework.
- Number of professor choices: 3.
- Number of framework choices per professor: 2. Total special project options = $3 \times 2 = 6$ .

Since the student chooses a chemistry project OR a physics project OR a special project, we use the Addition Principle to find the total number of options: Total options = (Chemistry options) + (Physics options) + (Special options) Total options =

8 + 5 + 6 = 19

. Distractor Rationale:

A is the result of adding all the numbers presented in the problem ( $8 + 5 + 3 + 2 = 18$ ), failing to use the multiplication principle for the special projects.
C is the result of incorrectly multiplying the first two categories and adding the third ( $8 \times 5 + (3 \times 2) = 40 + 6 = 46$ ).
D is the result of incorrectly multiplying all three categories of choices ( $8 \times 5 \times (3 \times 2) = 40 \times 6 = 240$ ), confusing 'or' choices with 'and' choices.