Inclusion-Exclusion Principle - Finite Mathematics

1. How many employees have both dental and vision insurance?
   Answer: 25 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.
2. How many integers from 1 to 500, inclusive, are divisible by 2, 3, or 5?
   Answer: 366 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.
3. Based on the survey results, how many students use exactly two of the three social media platforms?
   Answer: 90 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.
4. Based on the provided information, what is the value of n(A \cap B \cap C)?
   Answer: 1 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.
5. What percentage of students both play a sport and are in the school band?
   Answer: 10% 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%.
6. How many programmers know Java and Python, but NOT C++?
   Answer: 14 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.
7. How many of the logicians have at least one of these three features?
   Answer: 85 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.
8. How many of the athletes surveyed had tried none of the three supplements?
   Answer: 27 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.
9. How many integers from 1 to 1000, inclusive, are divisible by 6 or 10?
   Answer: 233 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.
10. How many students are taking Calculus but NOT Physics?
    Answer: 15 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.
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?
    Answer: $\frac{9}{20}$ of active patrons used exactly one service 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.
12. How many of the surveyed homeowners subscribe to exactly one of these three services?
    Answer: 35 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.
13. 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?
    Answer: 14 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.

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