# 1.1 MCQs-Sets

## Sets

### Basic Concepts and Definitions

1\. A set is defined as:

1. A collection of similar objects
2. A well-defined collection of distinct objects
3. A group of numbers
4. Any collection of things

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**Answer:** 2. A well-defined collection of distinct objects

**Explanation:**

* A set must be **well-defined**, meaning it's clear whether any given object is a member of the set or not.
* The objects (elements) in a set must be **distinct** (no repetitions).
* The order of elements does not matter in a set.
* Example: $${1, 2, 3}$$ and $${3, 1, 2}$$ represent the same set.

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2\. Which of the following is NOT a correct way to describe a set?

1. $${1, 3, 5, 7}$$
2. $${x \mid x \text{ is a prime number less than 10}}$$
3. $${2, 4, 6, 2, 8}$$
4. $${a, e, i, o, u}$$

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**Answer:** 3. $${2, 4, 6, 2, 8}$$

**Explanation:**

* In a set, elements should be **distinct** and not repeated.
* Option 3 has the element '2' repeated, which violates the definition of a set.
* Option 1 uses the **roster method** (listing all elements).
* Option 2 uses the **set-builder notation**.
* Option 4 is the set of vowels in English.

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3\. The set $${x : x \in \mathbb{N} \text{ and } x < 1}$$ is:

1. $${0}$$
2. $${1}$$
3. $$\emptyset$$
4. $${0, 1}$$

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**Answer:** 3. $$\emptyset$$

**Explanation:**

* The set contains all natural numbers ($$\mathbb{N}$$) that are less than 1.
* By convention, $$\mathbb{N} = {1, 2, 3, \ldots}$$ (natural numbers start from 1).
* There is **no natural number less than 1**.
* Therefore, the set is **empty**, denoted by $$\emptyset$$ or $${}$$.

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4\. Which of the following represents a singleton set?

1. $${a, b, c}$$
2. $${5}$$
3. $${\emptyset}$$
4. Both 2 and 3

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**Answer:** 4. Both 2 and 3

**Explanation:**

* A **singleton set** contains exactly one element.
* $${5}$$ has one element: the number 5.
* $${\emptyset}$$ has one element: the empty set $$\emptyset$$ itself.
* It's important to distinguish $$\emptyset$$ (empty set, 0 elements) from $${\emptyset}$$ (singleton set, 1 element).
* $${a, b, c}$$ has three elements, so it's not a singleton.

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5\. Two sets A and B are said to be equal if:

1. They have the same number of elements
2. Every element of A is in B and every element of B is in A
3. They are both subsets of each other
4. Both 2 and 3

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**Answer:** 4. Both 2 and 3

**Explanation:**

* Sets A and B are **equal** ($$A = B$$) if they contain exactly the same elements.
* Formally: $$A = B$$ if and only if $$A \subseteq B$$ and $$B \subseteq A$$.
* Option 2 is the direct definition: every element of A is in B, and every element of B is in A.
* Option 3 is equivalent to option 2 through the definition of subset.
* Option 1 describes **equivalent sets** (same cardinality), not necessarily equal sets.

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### Subsets and Power Sets

6\. If a set A has 5 elements, how many subsets does it have?

1. 5
2. 10
3. 25
4. 32

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**Answer:** 4. 32

**Explanation:**

* The number of subsets of a set with n elements is $$2^n$$.
* Here, n = 5, so number of subsets = $$2^5 = 32$$.
* This includes:
  * The empty set: 1
  * Singleton subsets: 5
  * 2-element subsets: $$\binom{5}{2} = 10$$
  * 3-element subsets: $$\binom{5}{3} = 10$$
  * 4-element subsets: $$\binom{5}{4} = 5$$
  * The set itself: 1
* Total = 1 + 5 + 10 + 10 + 5 + 1 = 32.

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7\. For any set A, which of the following is always TRUE?

1. $$A \in A$$
2. $$\emptyset \in A$$
3. $$\emptyset \subseteq A$$
4. $$A \subset A$$

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**Answer:** 3. $$\emptyset \subseteq A$$

**Explanation:**

* The **empty set** $$\emptyset$$ is a **subset of every set**, including itself.
* Formally: $$\emptyset \subseteq A$$ for any set A.
* $$A \in A$$ is generally false (except in some axiomatic set theories with non-well-founded sets).
* $$\emptyset \in A$$ is only true if A explicitly contains the empty set as an element.
* $$A \subset A$$ is false because $$A \subseteq A$$ (A is a subset of itself), but not a **proper** subset.

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8\. The power set of $${a, b}$$ is:

1. $${{a}, {b}}$$
2. $${\emptyset, {a}, {b}, {a, b}}$$
3. $${\emptyset, a, b, {a, b}}$$
4. $${a, b, {a}, {b}}$$

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**Answer:** 2. $${\emptyset, {a}, {b}, {a, b}}$$

**Explanation:**

* The **power set** of a set A, denoted $$P(A)$$, is the set of **all subsets** of A.
* For A = $${a, b}$$, the subsets are:
  * Empty set: $$\emptyset$$
  * Singleton subsets: $${a}$$, $${b}$$
  * The set itself: $${a, b}$$
* Therefore, $$P({a, b}) = {\emptyset, {a}, {b}, {a, b}}$$.
* Note: The elements of the power set are **sets themselves**.

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9\. If $$A \subseteq B$$ and $$B \subseteq A$$, then:

1. A and B are disjoint
2. A is a proper subset of B
3. A = B
4. B is a proper subset of A

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**Answer:** 3. A = B

**Explanation:**

* If $$A \subseteq B$$, then every element of A is in B.
* If $$B \subseteq A$$, then every element of B is in A.
* Together, these imply that A and B have exactly the same elements.
* Therefore, $$A = B$$.
* This is known as the **antisymmetry property** of set inclusion.

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10\. How many proper subsets does a set with 4 elements have?

1. 15
2. 16
3. 4
4. 8

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**Answer:** 1. 15

**Explanation:**

* For a set with n elements:
  * Total number of subsets = $$2^n$$
  * Number of **proper subsets** = $$2^n - 1$$ (excluding the set itself)
* Here, n = 4:
  * Total subsets = $$2^4 = 16$$
  * Proper subsets = 16 - 1 = 15
* Proper subsets include all subsets except the original set itself.

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### Universal Set and Complement

11\. If U is the universal set and A is any subset of U, then $$A \cup A'$$ equals:

1. $$\emptyset$$
2. A
3. $$A'$$
4. U

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**Answer:** 4. U

**Explanation:**

* $$A'$$ (complement of A) contains all elements in U that are **not** in A.
* The union $$A \cup A'$$ contains all elements that are either in A OR in A' (or both).
* Since every element of U is either in A or in A' (by definition of complement), the union equals the entire universal set.
* This is one of the **complement laws**: $$A \cup A' = U$$.

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12\. For any set A, $$(A')'$$ equals:

1. $$\emptyset$$
2. U
3. A
4. $$A'$$

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**Answer:** 3. A

**Explanation:**

* This is the **involution law** or **double complement law**.
* The complement of A, denoted $$A'$$, contains all elements not in A.
* The complement of $$A'$$, denoted $$(A')'$$, contains all elements not in $$A'$$.
* An element not in $$A'$$ must be in A.
* Therefore, $$(A')' = A$$.
* Example: If U = {1,2,3,4,5} and A = {1,2}, then A' = {3,4,5}, and (A')' = {1,2} = A.

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13\. Which of the following is NOT true for any sets A and B?

1. $$(A \cup B)' = A' \cap B'$$
2. $$(A \cap B)' = A' \cup B'$$
3. $$A - B = A \cap B'$$
4. $$A - B = B - A$$

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**Answer:** 4. $$A - B = B - A$$

**Explanation:**

* Options 1 and 2 are **De Morgan's Laws**.
* Option 3 is true: $$A - B$$ (elements in A but not in B) equals $$A \cap B'$$.
* Option 4 is **generally false**: $$A - B$$ and $$B - A$$ are different sets unless A = B.
* Example: Let A = {1,2,3}, B = {3,4,5}
  * $$A - B = {1,2}$$
  * $$B - A = {4,5}$$
  * Clearly, $${1,2} \neq {4,5}$$

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14\. If U = {1,2,3,4,5,6,7,8,9,10}, A = {2,4,6,8,10}, then $$A'$$ is:

1. {1,3,5,7,9}
2. {2,4,6,8,10}
3. {1,2,3,4,5,6,7,8,9,10}
4. $$\emptyset$$

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**Answer:** 1. {1,3,5,7,9}

**Explanation:**

* The complement $$A'$$ contains all elements in U that are **not** in A.
* U = {1,2,3,4,5,6,7,8,9,10}
* A = {2,4,6,8,10} (even numbers from 1 to 10)
* Elements not in A: {1,3,5,7,9} (odd numbers from 1 to 10)
* Therefore, $$A' = {1,3,5,7,9}$$.

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### Operations on Sets

15\. For sets A = {1,2,3,4} and B = {3,4,5,6}, $$A \cup B$$ is:

1. {3,4}
2. {1,2,5,6}
3. {1,2,3,4,5,6}
4. {1,2,3,4}

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**Answer:** 3. {1,2,3,4,5,6}

**Explanation:**

* The **union** $$A \cup B$$ contains all elements that are in A OR in B (or both).
* A = {1,2,3,4}
* B = {3,4,5,6}
* Elements in A: 1,2,3,4
* Elements in B: 3,4,5,6
* Combined unique elements: 1,2,3,4,5,6
* Therefore, $$A \cup B = {1,2,3,4,5,6}$$.

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16\. For sets A = {1,2,3,4} and B = {3,4,5,6}, $$A \cap B$$ is:

1. {3,4}
2. {1,2,5,6}
3. {1,2,3,4,5,6}
4. {1,2,3,4}

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**Answer:** 1. {3,4}

**Explanation:**

* The **intersection** $$A \cap B$$ contains all elements that are in BOTH A AND B.
* A = {1,2,3,4}
* B = {3,4,5,6}
* Common elements: 3 and 4
* Therefore, $$A \cap B = {3,4}$$.

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17\. For sets A = {1,2,3,4} and B = {3,4,5,6}, $$A - B$$ is:

1. {3,4}
2. {1,2}
3. {5,6}
4. {1,2,5,6}

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**Answer:** 2. {1,2}

**Explanation:**

* The **difference** $$A - B$$ contains elements that are in A but NOT in B.
* A = {1,2,3,4}
* B = {3,4,5,6}
* Elements in A: 1,2,3,4
* Remove elements that are also in B: 3,4
* Remaining: 1,2
* Therefore, $$A - B = {1,2}$$.

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18\. For sets A = {1,2,3,4} and B = {3,4,5,6}, the symmetric difference $$A \triangle B$$ is:

1. {3,4}
2. {1,2,5,6}
3. {1,2,3,4,5,6}
4. {1,2}

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**Answer:** 2. {1,2,5,6}

**Explanation:**

* **Symmetric difference** $$A \triangle B$$ contains elements that are in either A or B, but NOT in both.
* Definition: $$A \triangle B = (A - B) \cup (B - A)$$
* From previous questions:
  * $$A - B = {1,2}$$
  * $$B - A = {5,6}$$
* Therefore, $$A \triangle B = {1,2} \cup {5,6} = {1,2,5,6}$$.
* Alternatively: $$A \triangle B = (A \cup B) - (A \cap B) = {1,2,3,4,5,6} - {3,4} = {1,2,5,6}$$.

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### Laws of Set Algebra

19\. Which law states that $$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$?

1. Commutative law
2. Associative law
3. Distributive law
4. De Morgan's law

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**Answer:** 3. Distributive law

**Explanation:**

* This is the **distributive law of union over intersection**.
* Just like in algebra: $$a \times (b + c) = (a \times b) + (a \times c)$$
* In set theory: union distributes over intersection.
* There's also the dual distributive law: $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$
* This law can be proven using element arguments or Venn diagrams.

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20\. De Morgan's Law states that:

1. $$(A \cup B)' = A' \cap B'$$
2. $$(A \cap B)' = A' \cup B'$$
3. Both 1 and 2
4. $$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$

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**Answer:** 3. Both 1 and 2

**Explanation:**

* **De Morgan's Laws** are two important laws in set theory:
  1. The complement of a union equals the intersection of complements: $$(A \cup B)' = A' \cap B'$$
  2. The complement of an intersection equals the union of complements: $$(A \cap B)' = A' \cup B'$$
* These laws also apply in logic (for AND/OR operations) and Boolean algebra.
* They can be proven using element arguments or Venn diagrams.

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21\. The commutative law for sets states that:

1. $$A \cup B = B \cup A$$
2. $$A \cap B = B \cap A$$
3. Both 1 and 2
4. $$A \cup (B \cup C) = (A \cup B) \cup C$$

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**Answer:** 3. Both 1 and 2

**Explanation:**

* **Commutative laws** state that the order of operands doesn't matter for union and intersection:
  * $$A \cup B = B \cup A$$
  * $$A \cap B = B \cap A$$
* However, note that:
  * $$A - B \neq B - A$$ (difference is not commutative)
  * $$A \times B \neq B \times A$$ (Cartesian product is not commutative unless A = B)
* Option 4 represents the **associative law** for union.

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22\. Which of the following represents the absorption law?

1. $$A \cup (A \cap B) = A$$
2. $$A \cap (A \cup B) = A$$
3. Both 1 and 2
4. $$A \cup A = A$$

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**Answer:** 3. Both 1 and 2

**Explanation:**

* **Absorption laws** state that:
  1. $$A \cup (A \cap B) = A$$
  2. $$A \cap (A \cup B) = A$$
* These laws describe how a set "absorbs" certain combinations with itself.
* They can be verified using Venn diagrams or logical reasoning:
  * For (1): Any element in $$A \cup (A \cap B)$$ is either in A or in both A and B, so it must be in A.
  * For (2): Any element in $$A \cap (A \cup B)$$ is in A and also in either A or B, so it must be in A.
* Option 4 represents the **idempotent law**.

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### Cardinality of Sets

23\. For any two finite sets A and B, the formula $$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$ is known as:

1. De Morgan's principle
2. Inclusion-Exclusion principle
3. Distributive principle
4. Complement principle

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**Answer:** 2. Inclusion-Exclusion principle

**Explanation:**

* The **Inclusion-Exclusion principle** for two sets states: $$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$
* Reason: When we add $$n(A)$$ and $$n(B)$$, elements in $$A \cap B$$ are counted twice.
* So we subtract $$n(A \cap B)$$ once to count them only once.
* For three sets: $$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)$$

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24\. If $$n(A) = 20$$, $$n(B) = 30$$, and $$n(A \cap B) = 10$$, then $$n(A \cup B)$$ is:

1. 40
2. 50
3. 60
4. 70

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**Answer:** 1. 40

**Explanation:**

* Using the Inclusion-Exclusion principle: $$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$
* Substitute given values: $$n(A \cup B) = 20 + 30 - 10 = 40$$
* Verification:
  * Elements only in A: $$n(A) - n(A \cap B) = 20 - 10 = 10$$
  * Elements only in B: $$n(B) - n(A \cap B) = 30 - 10 = 20$$
  * Elements in both: $$n(A \cap B) = 10$$
  * Total: 10 + 20 + 10 = 40

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25\. If $$n(A) = 50$$, $$n(B) = 40$$, and $$n(A \cup B) = 70$$, then $$n(A \cap B)$$ is:

1. 10
2. 20
3. 30
4. 40

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**Answer:** 2. 20

**Explanation:**

* From Inclusion-Exclusion: $$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$
* Rearranging: $$n(A \cap B) = n(A) + n(B) - n(A \cup B)$$
* Substitute: $$n(A \cap B) = 50 + 40 - 70 = 20$$
* This means there are 20 elements common to both sets A and B.

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26\. In a class of 60 students, 35 play cricket and 25 play football. If 10 play both, how many play neither?

1. 10
2. 15
3. 20
4. 25

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**Answer:** 1. 10

**Explanation:**

* Let C = set of cricket players, F = set of football players
* Given: $$n(U) = 60$$, $$n(C) = 35$$, $$n(F) = 25$$, $$n(C \cap F) = 10$$
* Using Inclusion-Exclusion: $$n(C \cup F) = 35 + 25 - 10 = 50$$
* This means 50 students play at least one sport.
* Students playing neither = Total - Students playing at least one $$= 60 - 50 = 10$$

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### Cartesian Product

27\. The Cartesian product $$A \times B$$ is defined as:

1. $${a, b : a \in A, b \in B}$$
2. $${(a, b) : a \in A, b \in B}$$
3. $${b, a : b \in B, a \in A}$$
4. $$(a, b)$$ for all $$a \in A, b \in B$$

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**Answer:** 2. $${(a, b) : a \in A, b \in B}$$

**Explanation:**

* The **Cartesian product** $$A \times B$$ is the set of all **ordered pairs** $$(a, b)$$ where $$a \in A$$ and $$b \in B$$.
* Important properties:
  * $$(a, b)$$ is an ordered pair: $$(a, b) \neq (b, a)$$ unless a = b
  * $$A \times B \neq B \times A$$ in general
* Example: If A = {1,2}, B = {a,b}, then: $$A \times B = {(1,a), (1,b), (2,a), (2,b)}$$ $$B \times A = {(a,1), (a,2), (b,1), (b,2)}$$

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28\. If A has 3 elements and B has 4 elements, then $$A \times B$$ has:

1. 7 elements
2. 12 elements
3. 1 element
4. Cannot be determined

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**Answer:** 2. 12 elements

**Explanation:**

* The number of elements in $$A \times B$$ is the product of the number of elements in A and B.
* Formula: $$n(A \times B) = n(A) \times n(B)$$
* Here: $$n(A) = 3$$, $$n(B) = 4$$
* Therefore: $$n(A \times B) = 3 \times 4 = 12$$
* Each element of A can be paired with each element of B.

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29\. For any set A, $$A \times \emptyset$$ equals:

1. A
2. $$\emptyset$$
3. $${(a, \emptyset) : a \in A}$$
4. $${(\emptyset, a) : a \in A}$$

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**Answer:** 2. $$\emptyset$$

**Explanation:**

* The Cartesian product with the empty set is always empty.
* Reason: For $$A \times \emptyset$$ to have an element $$(a, b)$$, we need $$a \in A$$ and $$b \in \emptyset$$.
* But there is **no** $$b \in \emptyset$$ (empty set has no elements).
* Therefore, $$A \times \emptyset = \emptyset$$.
* Similarly, $$\emptyset \times A = \emptyset$$.

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30\. Which distributive property is TRUE for Cartesian product?

1. $$A \times (B \cup C) = (A \times B) \cup (A \times C)$$
2. $$A \times (B \cap C) = (A \times B) \cap (A \times C)$$
3. Both 1 and 2
4. Neither 1 nor 2

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**Answer:** 3. Both 1 and 2

**Explanation:**

* Cartesian product distributes over both union and intersection:
  1. $$A \times (B \cup C) = (A \times B) \cup (A \times C)$$
  2. $$A \times (B \cap C) = (A \times B) \cap (A \times C)$$
* Also distributes over set difference: $$A \times (B - C) = (A \times B) - (A \times C)$$
* These can be proven using element arguments: For (1): $$(x, y) \in A \times (B \cup C) \Leftrightarrow x \in A \text{ and } y \in (B \cup C)$$ $$\Leftrightarrow x \in A \text{ and } (y \in B \text{ or } y \in C)$$ $$\Leftrightarrow (x \in A \text{ and } y \in B) \text{ or } (x \in A \text{ and } y \in C)$$ $$\Leftrightarrow (x, y) \in (A \times B) \text{ or } (x, y) \in (A \times C)$$ $$\Leftrightarrow (x, y) \in (A \times B) \cup (A \times C)$$

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