Finite Mathematics Quiz: Inclusion Exclusion Principle
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Inclusion Exclusion PrincipleQuestion 1 of 13

In a group of 50 students, 30 are taking Calculus and 25 are taking Physics. It is also known that 10 students in this group are taking neither Calculus nor Physics.

How many students are taking Calculus but NOT Physics?

5
10
15
25
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Finite Mathematics Quiz

Finite Mathematics Quiz: Inclusion Exclusion Principle

Practice Inclusion Exclusion Principle 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 Inclusion Exclusion Principle, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

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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.

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Question 1

In a group of 50 students, 30 are taking Calculus and 25 are taking Physics. It is also known that 10 students in this group are taking neither Calculus nor Physics.

How many students are taking Calculus but NOT Physics?

  1. 5
  2. 10
  3. 15 (correct answer)
  4. 25
Explanation: Let C be the set of students taking Calculus and P be the set of students taking Physics. The total number of students is 50. Since 10 take neither, the number of students taking at least one of the courses is |C \cup P| = 50 - 10 = 40. We use the formula |C \cup P| = |C| + |P| - |C \cap P| to find the number of students taking both courses. 40 = 30 + 25 - |C \cap P|, which gives 40 = 55 - |C \cap P|, so |C \cap P| = 15. The number of students taking Calculus but not Physics is |C| - |C \cap P| = 30 - 15 = 15.

Question 2

A survey of 120 homeowners was conducted about their subscriptions to three services: Cable (C), Internet (I), and Landline (L). The survey showed: |C|=60, |I|=75, |L|=50, |C \cap I|=35, |C \cap L|=25, |I \cap L|=30, and |C \cap I \cap L|=10.

How many of the surveyed homeowners subscribe to exactly one of these three services?

  1. 35 (correct answer)
  2. 60
  3. 85
  4. 105
Explanation: The number of homeowners with exactly one service can be found by summing the number of people who have only that service. Only C = |C| - |C \cap I| - |C \cap L| + |C \cap I \cap L| = 60 - 35 - 25 + 10 = 10. Only I = |I| - |C \cap I| - |I \cap L| + |C \cap I \cap L| = 75 - 35 - 30 + 10 = 20. Only L = |L| - |C \cap L| - |I \cap L| + |C \cap I \cap L| = 50 - 25 - 30 + 10 = 5. The total number of homeowners with exactly one service is 10 + 20 + 5 = 35. Alternatively, use the formula (|C|+|I|+|L|) - 2(|C∩I|+|C∩L|+|I∩L|) + 3|C∩I∩L| = (60+75+50) - 2(35+25+30) + 3(10) = 185 - 2(90) + 30 = 185 - 180 + 30 = 35.

Question 3

Let A, B, and C be three finite sets, and let n(X) denote the cardinality of set X. You are given the following information:

  • n(A \ (B \cup C)) = 10
  • n(B \ (A \cup C)) = 15
  • n(C \ (A \cup B)) = 20
  • n(A \cap B) = 12
  • n(A \cap C) = 8
  • n(B \cap C) = 10
  • n(A \cup B \cup C) = 73

Based on the provided information, what is the value of n(A \cap B \cap C)?

  1. 1 (correct answer)
  2. 2
  3. 5
  4. 8
Explanation: Let x = n(A \cap B \cap C). The total union can be decomposed into seven disjoint regions. The terms n(A \ (B \cup C)), n(B \ (A \cup C)), and n(C \ (A \cup B)) represent the elements in exactly one of the sets. The number of elements in exactly two sets are n((A \cap B) \setminus C) = n(A \cap B) - x = 12 - x, n((A \cap C) \setminus B) = n(A \cap C) - x = 8 - x, and n((B \cap C) \setminus A) = n(B \cap C) - x = 10 - x. The union is the sum of these disjoint parts: n(A \cup B \cup C) = (n(A$\text{ only}$) + n(B$\text{ only}$) + n(C$\text{ only}$)) + (n(A,B$\text{ only}$) + n(A,C$\text{ only}$) + n(B,C$\text{ only}$)) + n(A,B,C). So, 73 = (10 + 15 + 20) + (12 - x) + (8 - x) + (10 - x) + x. This simplifies to 73 = 45 + 30 - 3x + x, which is 73 = 75 - 2x. Solving for x, we get 2x = 2, so x = 1.

Question 4

A software company has 80 programmers, each of whom knows at least one of the following three languages: Java, Python, or C++. The company's records show:

  • 50 know Java.
  • 42 know Python.
  • 30 know C++.
  • 15 know Java and C++.
  • 13 know Python and C++.
  • 10 know all three languages.

How many programmers know Java and Python, but NOT C++?

  1. 10
  2. 14 (correct answer)
  3. 24
  4. 34
Explanation: Let J, P, C be the sets of programmers who know Java, Python, and C++, respectively. We are given that |J \cup P \cup C| = 80. We want to find |J \cap P \cap C'|, which is equal to |J \cap P| - |J \cap P \cap C|. First, we must find |J \cap P|. We use the Inclusion-Exclusion Principle: |J \cup P \cup C| = |J| + |P| + |C| - (|J \cap P| + |J \cap C| + |P \cap C|) + |J \cap P \cap C|. Plugging in the known values: 80 = 50 + 42 + 30 - (|J \cap P| + 15 + 13) + 10. This simplifies to 80 = 122 - (|J \cap P| + 28) + 10, or 80 = 132 - |J \cap P| - 28, which gives 80 = 104 - |J \cap P|. Solving for |J \cap P| yields |J \cap P| = 24. Finally, the number who know Java and Python but not C++ is |J \cap P| - |J \cap P \cap C| = 24 - 10 = 14.

Question 5

At a certain high school, 60% of students play a sport, and 35% are in the school band. It is also known that 15% of students do neither activity.

What percentage of students both play a sport and are in the school band?

  1. 5%
  2. 10% (correct answer)
  3. 15%
  4. 25%
Explanation: Let S be the set of students who play a sport and B be the set of students in the band. We are given P(S) = 0.60, P(B) = 0.35, and P(S' \cap B') = 0.15. The event S' \cap B' is the complement of S \cup B. Therefore, P(S \cup B) = 1 - P((S \cup B)') = 1 - P(S' \cap B') = 1 - 0.15 = 0.85. Now, we use the Inclusion-Exclusion Principle for probability: P(S \cup B) = P(S) + P(B) - P(S \cap B). We have 0.85 = 0.60 + 0.35 - P(S \cap B), which simplifies to 0.85 = 0.95 - P(S \cap B). Solving for P(S \cap B) gives P(S \cap B) = 0.95 - 0.85 = 0.10, or 10%.

Question 6

A company has 150 employees. Each employee is offered optional insurance plans for health (H), dental (D), and vision (V). It is known that 90 employees have health insurance, 70 have dental, and 60 have vision. Additionally, 40 have health and dental, 35 have health and vision, and 20 have all three. A total of 10 employees have none of the three insurance plans.

How many employees have both dental and vision insurance?

  1. 5
  2. 15
  3. 25 (correct answer)
  4. 30
Explanation: Let n(X) be the number of employees with insurance plan X. The total number of employees is 150, and 10 have none, so the number of employees with at least one plan is n(H \cup D \cup V) = 150 - 10 = 140. Using the Inclusion-Exclusion Principle: n(H \cup D \cup V) = n(H) + n(D) + n(V) - (n(H \cap D) + n(H \cap V) + n(D \cap V)) + n(H \cap D \cap V). Let x = n(D \cap V). Then 140 = 90 + 70 + 60 - (40 + 35 + x) + 20. This simplifies to 140 = 220 - (75 + x) + 20, which is 140 = 240 - 75 - x, or 140 = 165 - x. Solving for x gives x = 165 - 140 = 25.

Question 7

A survey of 200 university students was conducted regarding their use of three social media platforms: Facebook, Twitter, and Instagram. The survey found that 120 students use Facebook, 90 use Twitter, and 110 use Instagram. Furthermore, 50 use Facebook and Twitter, 60 use Facebook and Instagram, 40 use Twitter and Instagram, and 20 students use all three platforms.

Based on the survey results, how many students use exactly two of the three social media platforms?

  1. 70
  2. 90 (correct answer)
  3. 110
  4. 150
Explanation: Let F, T, and I be the sets of students who use Facebook, Twitter, and Instagram, respectively. We want to find the number of students who use exactly two platforms. This is given by the formula: (|F \cap T| - |F \cap T \cap I|) + (|F \cap I| - |F \cap T \cap I|) + (|T \cap I| - |F \cap T \cap I|). This simplifies to |F \cap T| + |F \cap I| + |T \cap I| - 3|F \cap T \cap I|. Plugging in the given values: (50 + 60 + 40) - 3(20) = 150 - 60 = 90.

Question 8

A group of 100 athletes were surveyed about which of three performance-enhancing supplements (A, B, C) they had tried. The results were: 40 had tried A, 35 had tried B, and 30 had tried C. Furthermore, 15 had tried both A and B, 12 had tried both A and C, 10 had tried both B and C, and 5 had tried all three.

How many of the athletes surveyed had tried none of the three supplements?

  1. 27 (correct answer)
  2. 32
  3. 37
  4. 73
Explanation: First, we find the number of athletes who tried at least one supplement using the Inclusion-Exclusion Principle: |A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |A \cap C| + |B \cap C|) + |A \cap B \cap C|. Plugging in the values: |A \cup B \cup C| = (40 + 35 + 30) - (15 + 12 + 10) + 5 = 105 - 37 + 5 = 73. The number of athletes who tried none of the supplements is the total number of athletes minus those who tried at least one: 100 - 73 = 27.

Question 9

A manager needs to assign four different tasks (Task 1, Task 2, Task 3, Task 4) to four specific employees (A, B, C, D), with each employee receiving exactly one task.

If Employee A cannot be assigned Task 1 and Employee B cannot be assigned Task 2, in how many ways can the tasks be assigned?

  1. 10
  2. 12
  3. 14 (correct answer)
  4. 18
Explanation: The total number of ways to assign the tasks without any restrictions is 4! = 24. Let P_A be the set of assignments where Employee A gets Task 1, and P_B be the set of assignments where Employee B gets Task 2. We want to find the number of assignments in neither P_A nor P_B, which is |U| - |P_A \cup P_B|. If A gets Task 1, the other 3 employees can be assigned the remaining 3 tasks in 3! = 6 ways. So, |P_A| = 6. Similarly, if B gets Task 2, |P_B| = 6. If A gets Task 1 and B gets Task 2, the other 2 employees can be assigned the remaining 2 tasks in 2! = 2 ways. So, |P_A \cap P_B| = 2. By the Inclusion-Exclusion Principle, |P_A \cup P_B| = |P_A| + |P_B| - |P_A \cap P_B| = 6 + 6 - 2 = 10. The number of valid assignments is 24 - 10 = 14.

Question 10

At a conference of 100 logicians, some wear hats (H), some wear glasses (G), and some have beards (B). The following information is known:

  • 40 wear a hat.
  • 55 wear glasses.
  • 35 have a beard.
  • 20 wear a hat and glasses.
  • 15 wear a hat and have a beard.
  • 10 wear glasses and have a beard, but do not wear a hat.
  • 5 wear a hat, glasses, and have a beard.

How many of the logicians have at least one of these three features?

  1. 75
  2. 80
  3. 85 (correct answer)
  4. 90
Explanation: We want to find |H \cup G \cup B|. We are given |H|=40, |G|=55, |B|=35, |H \cap G|=20, |H \cap B|=15, and |H \cap G \cap B|=5. The piece of information |G \cap B \cap H'| = 10 means the number with glasses and a beard but no hat. To use the standard Inclusion-Exclusion formula, we need |G \cap B|. We know that |G \cap B| = |G \cap B \cap H'| + |G \cap B \cap H|. So, |G \cap B| = 10 + 5 = 15. Now we can apply the principle: |H \cup G \cup B| = |H|+|G|+|B| - (|H \cap G|+|H \cap B|+|G \cap B|) + |H \cap G \cap B|. |H \cup G \cup B| = 40+55+35 - (20+15+15) + 5 = 130 - 50 + 5 = 85.

Question 11

A library tracks patron usage of three digital services: e-books (E), audiobooks (A), and research databases (R). In a month with 500 active patrons, usage data shows: 250 used e-books, 200 used audiobooks, 150 used research databases, 100 used both e-books and audiobooks, 75 used both e-books and databases, 50 used both audiobooks and databases, and 25 used all three services. What fraction of active patrons used exactly one service?

  1. 720\frac{7}{20} of active patrons used exactly one service
  2. 25\frac{2}{5} of active patrons used exactly one service
  3. 920\frac{9}{20} of active patrons used exactly one service (correct answer)
  4. 12\frac{1}{2} of active patrons used exactly one service
Explanation: First, find patrons using exactly one service: E only = 250 - 100 - 75 + 25 = 100, A only = 200 - 100 - 50 + 25 = 75, R only = 150 - 75 - 50 + 25 = 50. Total exactly one = 100 + 75 + 50 = 225. Fraction = 225/500 = 9/20. Choice A (175/500 = 7/20) represents a common error of not adding back the triple intersection. Choice B (200/500 = 2/5) miscalculates overlaps. Choice D (250/500 = 1/2) ignores inclusion-exclusion entirely.

Question 12

How many integers from 1 to 500, inclusive, are divisible by 2, 3, or 5?

  1. 334
  2. 350
  3. 366 (correct answer)
  4. 516
Explanation: Let A_k be the set of integers from 1 to 500 divisible by k. We want |A_2 \cup A_3 \cup A_5|. |A_2| = \lfloor 500/2 \rfloor = 250. |A_3| = \lfloor 500/3 \rfloor = 166. |A_5| = \lfloor 500/5 \rfloor = 100. |A_2 \cap A_3| = |A_6| = \lfloor 500/6 \rfloor = 83. |A_2 \cap A_5| = |A_{10}| = \lfloor 500/10 \rfloor = 50. |A_3 \cap A_5| = |A_{15}| = \lfloor 500/15 \rfloor = 33. |A_2 \cap A_3 \cap A_5| = |A_{30}| = \lfloor 500/30 \rfloor = 16. By the Inclusion-Exclusion Principle: (250+166+100) - (83+50+33) + 16 = 516 - 166 + 16 = 366.

Question 13

How many integers from 1 to 1000, inclusive, are divisible by 6 or 10?

  1. 233 (correct answer)
  2. 250
  3. 266
  4. 300
Explanation: Let A be the set of integers from 1 to 1000 divisible by 6, and B be the set of integers divisible by 10. We want to find |A \cup B| = |A| + |B| - |A \cap B|. The number of integers divisible by k is $\lfloor 1000/k \rfloor$. So, |A| = \lfloor 1000/6 \rfloor = 166, and |B| = \lfloor 1000/10 \rfloor = 100. The intersection A \cap B is the set of integers divisible by the least common multiple of 6 and 10, which is lcm(6, 10) = 30. Thus, |A \cap B| = \lfloor 1000/30 \rfloor = 33. Applying the principle, |A \cup B| = 166 + 100 - 33 = 233.