Set notation and operations (union, intersection, complement) - Discrete Math

Let $U = \{1, 2, 3, ..., 20\}$ and consider sets $M = \{x \in U : x \text{ is a multiple of } 3\}$ and $N = \{x \in U : x \text{ is a multiple of } 4\}$ . If $T = M^c \cup N^c$ , which element is definitely NOT in $T$ ?
Answer: $12$ (since it's a multiple of both 3 and 4) We have M = {3, 6, 9, 12, 15, 18} and N = {4, 8, 12, 16, 20}. By De Morgan's law, T = $M^c$ ∪ $N^c$ = $(M ∩ N)^c$ . So T contains all elements except those in M ∩ N = {12}. Therefore, 12 is the only element definitely NOT in T. Choice A (11) is in T since 11 ∉ M and 11 ∉ N, so 11 ∈ $M^c$ and 11 ∈ $N^c$ . Choice C (15) is in T since 15 ∉ N, so 15 ∈ $N^c$ . Choice D (16) is in T since 16 ∉ M, so 16 ∈ $M^c$ .
Consider sets $A$ , $B$ , and $C$ such that $A \cup B \cup C = U$ , $|A| = |B| = |C| = 10$ , $|A \cap B| = |B \cap C| = |A \cap C| = 4$ , and $|A \cap B \cap C| = 2$ . Using the principle of inclusion-exclusion, what is $|U|$ ?
Answer: $20$ When you encounter problems involving multiple overlapping sets, the principle of inclusion-exclusion is your key tool. This principle systematically counts elements while avoiding double-counting those that appear in multiple sets. For three sets, the inclusion-exclusion formula is: $|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|$ Let's substitute the given values: $|U| = 10 + 10 + 10 - 4 - 4 - 4 + 2 = 30 - 12 + 2 = 20$ The logic here is that we first add all individual set sizes (30 total), then subtract the pairwise intersections because those elements were counted twice (subtracting 12), and finally add back the three-way intersection because it was subtracted three times but should only be subtracted twice (adding back 2). Looking at the wrong answers: Choice A (16) likely results from forgetting to add back the three-way intersection: $30 - 12 = 18$ , then making an arithmetic error. Choice C (18) is exactly what you get if you forget the final step of adding back $|A \cap B \cap C|$ . Choice D (22) might come from incorrectly adding the three-way intersection twice or making sign errors in the calculation. The correct answer is B) 20. Study tip: Always write out the inclusion-exclusion formula completely before substituting values. The alternating pattern of signs (add individual sets, subtract pairwise intersections, add three-way intersection) is crucial and easy to mess up if you try to do it mentally.
If $|A \triangle B| = 15$ and $|A \cap B| = 8$ , where $\triangle$ denotes symmetric difference, what is $|A \cup B|$ ?
Answer: $23$ elements The symmetric difference A △ B = (A \ B) ∪ (B \ A), so |A △ B| = |A \ B| + |B \ A| = 15. Since A ∪ B can be partitioned into three disjoint parts: (A \ B), (A ∩ B), and (B \ A), we have |A ∪ B| = |A \ B| + |A ∩ B| + |B \ A| = |A △ B| + |A ∩ B| = 15 + 8 = 23. Choice A incorrectly uses only |A △ B|. Choice C incorrectly doubles |A △ B|. Choice D incorrectly adds |A △ B| + |A ∩ B| + |A △ B|.
For finite sets $P$ and $Q$ , suppose $|P \cup Q| = 25$ , $|P \cap Q| = 8$ , and $|P^c \cap Q| = 12$ . What is $|P \setminus Q|$ ?
Answer: $5$ elements When you encounter set problems with unions, intersections, and complements, visualizing with a Venn diagram helps you organize the given information and find relationships between different regions. Let's work systematically with what we know. Since $|P^c \cap Q| = 12$ , this tells us there are 12 elements that are in $Q$ but not in $P$ . We also know $|P \cap Q| = 8$ , meaning 8 elements are in both sets. Therefore, the total number of elements in $Q$ is $|Q| = |P^c \cap Q| + |P \cap Q| = 12 + 8 = 20$ . Now we can find $|P \setminus Q|$ using the union formula. Since $|P \cup Q| = |P| + |Q| - |P \cap Q|$ , we have $25 = |P| + 20 - 8$ , which gives us $|P| = 13$ . The set $P \setminus Q$ contains elements that are in $P$ but not in $Q$ . Since $|P| = 13$ and $|P \cap Q| = 8$ , we get $|P \setminus Q| = |P| - |P \cap Q| = 13 - 8 = 5$ . Looking at the wrong answers: (A) 17 likely comes from incorrectly calculating $25 - 8 = 17$ , confusing the union size with something else. (B) 8 mistakenly uses the intersection size instead of the difference. (C) 13 uses $|P|$ directly, forgetting to subtract the intersection. The key strategy is to break down complex set relationships into their basic components using the fundamental counting principles. Always identify what each region of your mental Venn diagram represents before calculating.
For any sets $A$ and $B$ , define $A \oplus B = (A \setminus B) \cup (B \setminus A)$ . If $P \oplus Q = \{1, 2, 6, 7\}$ and $P \cap Q = \{3, 5\}$ , which of the following could be $P \cup Q$ ?
Answer: $\{1, 2, 3, 5, 6, 7\}$ When you encounter set operations like the symmetric difference $\oplus$ , focus on understanding what each operation tells you about the relationship between sets. The symmetric difference $A \oplus B = (A \setminus B) \cup (B \setminus A)$ gives you elements that are in exactly one of the two sets, but not in both. Since $P \oplus Q = \{1, 2, 6, 7\}$ , these elements belong to either $P$ or $Q$ , but not both. Since $P \cap Q = \{3, 5\}$ , these elements belong to both sets. To find $P \cup Q$ , you need all elements that belong to at least one of the sets, which means combining the symmetric difference with the intersection: $P \cup Q = (P \oplus Q) \cup (P \cap Q) = \{1, 2, 6, 7\} \cup \{3, 5\} = \{1, 2, 3, 5, 6, 7\}$ . Looking at the choices: Choice A $\{1, 3, 5, 7\}$ is missing elements 2 and 6 from the symmetric difference, so it's incomplete. Choice B $\{1, 2, 3, 4, 5, 6, 7\}$ includes element 4, which appears in neither the symmetric difference nor the intersection, making it impossible. Choice C $\{2, 3, 5, 6\}$ is missing elements 1 and 7 from the symmetric difference. Choice D $\{1, 2, 3, 5, 6, 7\}$ correctly includes all elements from both given sets. Remember this key relationship: $P \cup Q = (P \oplus Q) \cup (P \cap Q)$ . The union always equals the symmetric difference plus the intersection, with no extra elements.
Let $E = \{n \in \mathbb{Z}^+ : n \leq 50 \text{ and } n \text{ is even}\}$ and $F = \{n \in \mathbb{Z}^+ : n \leq 50 \text{ and } n \equiv 1 \pmod{3}\}$ . What is $|E \cap F|$ ?
Answer: $8$ elements Set E contains even numbers from 1 to 50: E = {2, 4, 6, 8, ..., 50}. Set F contains positive integers ≤ 50 that are ≡ 1 (mod 3): F = {1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49}. For E ∩ F, we need numbers that are both even and ≡ 1 (mod 3). These are: 4, 10, 16, 22, 28, 34, 40, 46. These form an arithmetic sequence with first term 4 and common difference 6. The general term is 4 + 6k where k ≥ 0. We need 4 + 6k ≤ 50, so k ≤ 46/6 = 7.67, giving k ∈ {0, 1, 2, 3, 4, 5, 6, 7}, which is 8 values. Choice A miscounts by missing one element. Choice C incorrectly includes an odd number. Choice D uses |F| instead of |E ∩ F|.
Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ , $A = \{x \in U : x \text{ is odd}\}$ , and $B = \{x \in U : x \leq 6\}$ . If $C = (A \cup B)^c \cap (A \cap B)$ , what is $|C|$ ?
Answer: $0$ First, find A = {1, 3, 5, 7, 9} and B = {1, 2, 3, 4, 5, 6}. Then A ∪ B = {1, 2, 3, 4, 5, 6, 7, 9}, so $(A ∪ B)^c$ = {8, 10}. Also, A ∩ B = {1, 3, 5}. Therefore, C = $(A ∪ B)^c$ ∩ (A ∩ B) = {8, 10} ∩ {1, 3, 5} = ∅. So |C| = 0. Choice B incorrectly uses | $(A ∪ B)^c$ | = 2. Choice C incorrectly uses |A ∩ B| = 3. Choice D incorrectly uses |A| = 5.
Let $S = \{a, b, c, d, e\}$ and consider the family of sets $\mathcal{F} = \{\{a, b\}, \{b, c, d\}, \{a, d, e\}, \{c, e\}\}$ . What is $\bigcap_{F \in \mathcal{F}} F^c$ , where $F^c$ denotes the complement of $F$ with respect to $S$ ?
Answer: $\emptyset$ First find complements: {a,b}^c = {c,d,e}, {b,c,d}^c = {a,e}, {a,d,e}^c = {b,c}, {c,e}^c = {a,b,d}. The intersection of all complements is {c,d,e} ∩ {a,e} ∩ {b,c} ∩ {a,b,d}. Since {a,e} ∩ {b,c} = ∅, the entire intersection is empty. By De Morgan's law, this equals (⋃F∈ℱ F)^c, and since every element of S appears in at least one set in ℱ, the union is S, making the complement empty. Choice A incorrectly intersects original sets. Choice B miscomputes complements. Choice D finds intersection of just two complements.
If $A \subseteq B \subseteq C$ and $|A| = 8$ , $|B \setminus A| = 5$ , $|C \setminus B| = 7$ , what is $|A^c \cap C|$ ?
Answer: $12$ elements When you encounter nested subset relationships with cardinality information, visualize the sets as concentric circles and work systematically through the regions created by the set operations. Given $A \subseteq B \subseteq C$ , you can think of $A$ inside $B$ , which is inside $C$ . The complement $A^c \cap C$ represents everything in $C$ that's not in $A$ . Since $A \subseteq C$ , this equals $C \setminus A$ . To find $|C \setminus A|$ , break it into two disjoint parts: elements in $B$ but not $A$ , and elements in $C$ but not $B$ . You're given $|B \setminus A| = 5$ and $|C \setminus B| = 7$ . Therefore: $|C \setminus A| = |B \setminus A| + |C \setminus B| = 5 + 7 = 12$ . Choice A (15) likely comes from incorrectly adding all three given numbers: $8 + 5 + 7 = 20$ , then subtracting $|A| = 8$ to get 12, but then making an arithmetic error. Choice B (13) might result from adding $|A| = 8$ and $|B \setminus A| = 5$ , confusing this with $|C \setminus A|$ . Choice D (20) comes from simply adding all three given cardinalities without understanding what the question asks for. Remember that when working with nested subsets and complements, always identify exactly which regions you need to count, then use the given information to build up to that total. Drawing a Venn diagram can prevent calculation errors and conceptual confusion.
In a survey of 100 students, let $S$ be the set of students who study Spanish, $F$ be the set who study French, and $G$ be the set who study German. If $|S \cup F \cup G| = 85$ , $|S \cap F \cap G| = 5$ , and $|(S \cup F \cup G)^c| = 15$ , what is the number of students who study none of these languages?
Answer: $15$ students The students who study none of these languages are exactly those in $(S ∪ F ∪ G)^c$ . We're given that | $(S ∪ F ∪ G)^c$ | = 15, so 15 students study none of these languages. This can also be verified: if |S ∪ F ∪ G| = 85 out of 100 total students, then 100 - 85 = 15 students are not in S ∪ F ∪ G. Choice A incorrectly uses |S ∩ F ∩ G|. Choice B uses half of the complement. Choice D adds |S ∩ F ∩ G| to | $(S ∪ F ∪ G)^c$ |.
Let $S = \{n \in \mathbb{Z} : -10 \leq n \leq 10\}$ and define $A = \{n \in S : n^2 - 3n \leq 0\}$ . If $B = \{n \in S : |n| \geq 4\}$ , what is $(A \cup B)^c \cap S$ ?
Answer: $\{-3, -2, -1\}$ First, find A by solving n² - 3n ≤ 0, which factors as n(n-3) ≤ 0. This gives 0 ≤ n ≤ 3, so A = {0, 1, 2, 3}. Next, B = {n ∈ S : |n| ≥ 4} = {-10, -9, ..., -4, 4, 5, ..., 10}. Then A ∪ B = {-10, -9, ..., -4, 0, 1, 2, 3, 4, 5, ..., 10}. The complement $(A ∪ B)^c$ ∩ S consists of elements in S but not in A ∪ B, which are {-3, -2, -1}. Choice B incorrectly includes 4 ∈ B. Choice C misses -3. Choice D gives A ∪ B ∩ {-3, -2, -1, 4} instead of the complement.
Let $\mathcal{A} = \{A_1, A_2, A_3, A_4\}$ where $A_1 = \{1, 2, 5\}$ , $A_2 = \{2, 3, 4\}$ , $A_3 = \{1, 4, 6\}$ , and $A_4 = \{3, 5, 6\}$ . Consider the set $M = \bigcup_{i=1}^{4} A_i$ and define $B = \{x \in M : x \text{ appears in exactly two sets from } \mathcal{A}\}$ . What is $|B|$ ?
Answer: $6$ First, M = A₁ ∪ A₂ ∪ A₃ ∪ A₄ = {1, 2, 3, 4, 5, 6}. Now count appearances: 1 appears in A₁, A₃ (exactly 2); 2 appears in A₁, A₂ (exactly 2); 3 appears in A₂, A₄ (exactly 2); 4 appears in A₂, A₃ (exactly 2); 5 appears in A₁, A₄ (exactly 2); 6 appears in A₃, A₄ (exactly 2). Therefore B = {1, 2, 3, 4, 5, 6} and |B| = 6.
For sets $X$ , $Y$ , and $Z$ , which of the following statements about $(X \triangle Y) \cap Z$ is always true, where $\triangle$ denotes symmetric difference?
Answer: It equals $(X \cap Z) \triangle (Y \cap Z)$ for any choice of sets The symmetric difference distributes over intersection: (X △ Y) ∩ Z = (X ∩ Z) △ (Y ∩ Z). This can be proven by showing both sides contain exactly the elements in Z that belong to exactly one of X or Y. Choice B confuses the operations. Choice C gives an incorrect formula even under the stated condition. Choice D is incorrect; when X ∩ Y ∩ Z = ∅, the expression still equals (X ∩ Z) △ (Y ∩ Z), not (X ∪ Y) ∩ Z.
Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ , $A = \{x \in U : x \text{ is odd}\}$ , $B = \{x \in U : x > 5\}$ , and $C = \{x \in U : x \text{ is divisible by } 3\}$ . What is $(A \cap B) \cup (\overline{C} \cap A)$ ?
Answer: $\{1, 5, 7, 9\}$ First, identify the sets: A = {1, 3, 5, 7, 9}, B = {6, 7, 8, 9, 10}, C = {3, 6, 9}. Then A ∩ B = {7, 9}. For the complement, C̄ = {1, 2, 4, 5, 7, 8, 10}, so C̄ ∩ A = {1, 5, 7}. Finally, (A ∩ B) ∪ (C̄ ∩ A) = {7, 9} ∪ {1, 5, 7} = {1, 5, 7, 9}. Choice A incorrectly includes 3 and omits 5. Choice B omits 5. Choice D omits 1.

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