Cartesian Products of Sets Questions in English

(A) The Cartesian product of sets follows the distributive law over the union of sets.
According to the distributive property of Cartesian product over union:
$A \times (B \cup C) = (A \times B) \cup (A \times C)$.
Therefore,the correct option is $A$.

(B) Given sets are $A = \{ 2, 4, 5 \}$ and $B = \{ 7, 8, 9 \}$.
Number of elements in set $A$ is $n(A) = 3$.
Number of elements in set $B$ is $n(B) = 3$.
The number of elements in the Cartesian product $A \times B$ is given by the formula $n(A \times B) = n(A) \times n(B)$.
Therefore,$n(A \times B) = 3 \times 3 = 9$.

(C) The Cartesian product of two sets $A$ and $B$,denoted by $A \times B$,is defined as the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$.
If the number of elements in set $A$ is $n(A) = p$ and the number of elements in set $B$ is $n(B) = q$,then the number of elements in the Cartesian product $A \times B$ is given by the formula:
$n(A \times B) = n(A) \times n(B) = p \times q = pq$.

(C) Given sets are $A = \{a, b\}$,$B = \{c, d\}$,and $C = \{d, e\}$.
First,find the union of sets $B$ and $C$:
$B \cup C = \{c, d\} \cup \{d, e\} = \{c, d, e\}$.
Now,calculate the Cartesian product $A \times (B \cup C)$:
$A \times (B \cup C) = \{a, b\} \times \{c, d, e\} = \{(a, c), (a, d), (a, e), (b, c), (b, d), (b, e)\}$.
Thus,the given set is equal to $A \times (B \cup C)$.

(C) The Cartesian product of two sets $A$ and $B$ is defined as $A \times B = \{(a, b) : a \in A, b \in B\}$.
In general,$A \times B \neq B \times A$ unless the sets are equal or one of them is empty.
Specifically,$A \times B = B \times A$ if and only if $A = B$ (assuming $A, B \neq \emptyset$).
Thus,the correct option is $C$.

(B) Given $A = \{1, 2, 4\}, B = \{2, 4, 5\}, C = \{2, 5\}$.
First,calculate $A - B$:
$A - B = \{x : x \in A \text{ and } x \notin B\} = \{1\}$.
Next,calculate $B - C$:
$B - C = \{x : x \in B \text{ and } x \notin C\} = \{4\}$.
Finally,the Cartesian product $(A - B) \times (B - C)$ is:
$\{1\} \times \{4\} = \{(1, 4)\}$.

(A) Given that $(1, 3), (2, 5), (3, 3) \in A \times B$.
From these ordered pairs,the elements of set $A$ are the first components: $A = \{1, 2, 3\}$.
The elements of set $B$ are the second components: $B = \{3, 5\}$.
The total number of elements in $A \times B$ is $n(A) \times n(B) = 3 \times 2 = 6$.
The complete set $A \times B$ is $\{(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5)\}$.
We are given three elements: $(1, 3), (2, 5), (3, 3)$.
The remaining elements are $\{(1, 5), (2, 3), (3, 5)\}$.
Thus,the correct option is $A$.

(B) Given $A = \{1, 2, 3\}$ and $B = \{3, 8\}$.
First,find the union of sets $A$ and $B$: $A \cup B = \{1, 2, 3, 8\}$.
Next,find the intersection of sets $A$ and $B$: $A \cap B = \{3\}$.
Now,calculate the Cartesian product $(A \cup B) \times (A \cap B)$:
$(A \cup B) \times (A \cap B) = \{1, 2, 3, 8\} \times \{3\} = \{(1, 3), (2, 3), (3, 3), (8, 3)\}$.

(C) Given sets are $A = \{2, 3, 5\}$ and $B = \{2, 5, 6\}$.
First,find the set difference $(A - B)$:
$A - B = \{x : x \in A \text{ and } x \notin B\} = \{3\}$.
Next,find the intersection $(A \cap B)$:
$A \cap B = \{x : x \in A \text{ and } x \in B\} = \{2, 5\}$.
Finally,calculate the Cartesian product $(A - B) \times (A \cap B)$:
$(A - B) \times (A \cap B) = \{3\} \times \{2, 5\} = \{(3, 2), (3, 5)\}$.

(C) The Cartesian product of two sets $A$ and $B$,denoted by $A \times B$,is the set of all ordered pairs $(x, y)$ such that $x \in A$ and $y \in B$.
Given $A = \{1, 2, 3, 4\}$ and $B = \{a, b\}$.
$A \times B = \{(1, a), (1, b), (2, a), (2, b), (3, a), (3, b), (4, a), (4, b)\}$.
Thus,the correct option is $C$.

(B) The number of elements common to $A \times B$ and $B \times A$ is given by $n((A \times B) \cap (B \times A))$.
Using the property of Cartesian products,$(A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A)$.
Therefore,$n((A \times B) \cap (B \times A)) = n((A \cap B) \times (A \cap B)) = n(A \cap B) \times n(A \cap B)$.
Given that $n(A \cap B) = 99$,we have $99 \times 99 = 99^2$.

(C) By definition,a relation $R$ from a set $A$ to a set $B$ is a subset of the Cartesian product $A \times B$.
Therefore,$R \subseteq A \times B$.
Since the question asks for the set of which $R$ is a subset,the correct option is $A \times B$.

(A) First,we solve the equation $x^2 - 5x + 6 = 0$ to find the elements of set $A$.
$x^2 - 2x - 3x + 6 = 0 \implies x(x - 2) - 3(x - 2) = 0 \implies (x - 2)(x - 3) = 0$.
Thus,$A = \{2, 3\}$.
Given $B = \{2, 4\}$ and $C = \{4, 5\}$,the intersection $B \cap C$ is the set of common elements.
$B \cap C = \{4\}$.
Now,we find the Cartesian product $A \times (B \cap C) = \{2, 3\} \times \{4\}$.
$A \times (B \cap C) = \{(2, 4), (3, 4)\}$.
Therefore,the correct option is $A$.

(C) The intersection of two Cartesian products is given by the formula $(A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A)$.
Given $A = \{1, 2, 3, 4, 5\}$ and $B = \{2, 3, 6, 7\}$.
The intersection of the sets is $A \cap B = \{2, 3\}$.
Thus,$(A \cap B) \times (B \cap A) = \{2, 3\} \times \{2, 3\} = \{(2, 2), (2, 3), (3, 2), (3, 3)\}$.
The number of elements in this set is $2 \times 2 = 4$.

(B) Let $n(A) = 4$ and $n(B) = 2$.
Then the number of elements in $A \times B$ is $n(A \times B) = 4 \times 2 = 8$.
The total number of subsets of $A \times B$ is $2^8 = 256$.
We need to find the number of subsets having at least $3$ elements.
This is equal to the total number of subsets minus the number of subsets having $0, 1,$ or $2$ elements.
Number of subsets with $0$ elements = $\binom{8}{0} = 1$.
Number of subsets with $1$ element = $\binom{8}{1} = 8$.
Number of subsets with $2$ elements = $\binom{8}{2} = \frac{8 \times 7}{2} = 28$.
Total number of subsets with less than $3$ elements = $1 + 8 + 28 = 37$.
Therefore,the number of subsets with at least $3$ elements = $256 - 37 = 219$.

(B) The Cartesian product of sets distributes over the intersection of sets.
Therefore,$A \times (B \cap C) = (A \times B) \cap (A \times C).$
This is known as the distributive law of Cartesian product over intersection.

(D) By the definition of the Cartesian product of two sets $A$ and $B$,$A \times B = \{(a, b) : a \in A, b \in B\}$.
Given $A = \{0, 1\}$ and $B = \{1, 0\}$.
We pair each element of $A$ with each element of $B$:
For $a = 0$: $(0, 1)$ and $(0, 0)$.
For $a = 1$: $(1, 1)$ and $(1, 0)$.
Therefore,$A \times B = \{(0, 1), (0, 0), (1, 1), (1, 0)\}$.

(C) The number of elements in the Cartesian product of two sets $A$ and $B$ is given by the product of the number of elements in each set.
Given $n(A) = p$ and $n(B) = q$.
Therefore,$n(A \times B) = n(A) \times n(B) = p \times q = pq$.

(C) Given sets are $A = \{a, b\}$,$B = \{c, d\}$,and $C = \{d, e\}$.
First,find the union of sets $B$ and $C$:
$B \cup C = \{c, d\} \cup \{d, e\} = \{c, d, e\}$.
Now,find the Cartesian product of $A$ and $(B \cup C)$:
$A \times (B \cup C) = \{a, b\} \times \{c, d, e\} = \{(a, c), (a, d), (a, e), (b, c), (b, d), (b, e)\}$.
Thus,the given set is equal to $A \times (B \cup C)$.

(A) Given $A = \{x : x^2 - 5x + 6 = 0\}$.
Solving the quadratic equation $x^2 - 5x + 6 = 0$,we get $(x - 2)(x - 3) = 0$,so $A = \{2, 3\}$.
Given $B = \{2, 4\}$ and $C = \{4, 5\}$.
The intersection $B \cap C$ is the set of common elements,so $B \cap C = \{4\}$.
Now,the Cartesian product $A \times (B \cap C) = \{2, 3\} \times \{4\}$.
This results in the set of ordered pairs $\{(2, 4), (3, 4)\}$.

(A) Given sets are $A = \{2, 3, 5\}$ and $B = \{2, 5, 6\}$.
First,find the difference set $(A - B)$:
$A - B$ consists of elements that are in $A$ but not in $B$.
$A - B = \{3\}$.
Next,find the intersection set $(A \cap B)$:
$A \cap B$ consists of elements common to both $A$ and $B$.
$A \cap B = \{2, 5\}$.
Finally,calculate the Cartesian product $(A - B) \times (A \cap B)$:
$(A - B) \times (A \cap B) = \{3\} \times \{2, 5\} = \{(3, 2), (3, 5)\}$.

(C) By definition,a relation $R$ from a set $A$ to a set $B$ is a subset of the Cartesian product $A \times B$.
Since $A = \{a, b, c\}$ and $B = \{1, 2\}$,the Cartesian product $A \times B$ is the set of all ordered pairs $(x, y)$ such that $x \in A$ and $y \in B$.
Thus,$A \times B = \{(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)\}$.
Any relation $R$ from $A$ to $B$ must satisfy $R \subseteq A \times B$.

(D) We know that for any sets $A, B, C$,the intersection of Cartesian products is given by the formula: $(A \times B) \cap (B \times C) = (A \cap B) \times (B \cap C)$.
Given that $n(A \cap B) = 2$ and $n(B \cap C) = 2$.
Therefore,the number of elements in the intersection is $n((A \cap B) \times (B \cap C)) = n(A \cap B) \times n(B \cap C)$.
Substituting the given values,we get $2 \times 2 = 4$.
Thus,$n((A \times B) \cap (B \times C)) = 4$.

(C) Given $n(A) = 3$ and $n(B) = 3$,the number of elements in the Cartesian product $A \times B$ is $n(A \times B) = n(A) \times n(B) = 3 \times 3 = 9$.
The number of subsets of a set with $m$ elements having an odd number of elements is given by the sum of combinations $\sum_{k \text{ is odd}} \binom{m}{k} = \binom{m}{1} + \binom{m}{3} + \dots + \binom{m}{m}$.
For $m = 9$,this sum is $\binom{9}{1} + \binom{9}{3} + \binom{9}{5} + \binom{9}{7} + \binom{9}{9}$.
Using the identity $\sum_{k \text{ is odd}} \binom{m}{k} = 2^{m-1}$,we get $2^{9-1} = 2^8 = 256$.

(C) The set $A$ contains elements from $1$ to $100$,so $n(A) = 100$.
The set $B$ contains elements from $51$ to $180$,so $n(B) = 180 - 51 + 1 = 130$.
The intersection $A \cap B$ contains elements from $51$ to $100$,so $n(A \cap B) = 100 - 51 + 1 = 50$.
We know that $(A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A)$.
Therefore,the number of elements is $n((A \cap B) \times (B \cap A)) = n(A \cap B) \times n(B \cap A)$.
Since $n(A \cap B) = 50$,we have $50 \times 50 = 2500$.

(A) Since the ordered pairs are equal,their corresponding components must be equal.
Therefore,we have:
$x + 1 = 3 \implies x = 3 - 1 = 2$
And,
$y - 2 = 1 \implies y = 1 + 2 = 3$
Thus,the values are $x = 2$ and $y = 3$.

(B) By the definition of the Cartesian product:
$P \times Q = \{(a, r), (b, r), (c, r)\}$
$Q \times P = \{(r, a), (r, b), (r, c)\}$
Since,by the definition of equality of ordered pairs,the pair $(a, r)$ is not equal to the pair $(r, a)$,we conclude that $P \times Q \neq Q \times P$.
However,the number of elements in each set is the same.

(A) First,find the intersection of sets $B$ and $C$:
$B \cap C = \{3, 4\} \cap \{4, 5, 6\} = \{4\}$.
Next,find the Cartesian product of set $A$ and the resulting set $\{4\}$:
$A \times (B \cap C) = \{1, 2, 3\} \times \{4\} = \{(1, 4), (2, 4), (3, 4)\}$.

(A) We know that $(A \times B) \cap (A \times C) = A \times (B \cap C)$.
First,find the intersection of sets $B$ and $C$:
$B \cap C = \{3, 4\} \cap \{4, 5, 6\} = \{4\}$.
Now,calculate the Cartesian product $A \times (B \cap C)$:
$A \times \{4\} = \{(1, 4), (2, 4), (3, 4)\}$.
Thus,$(A \times B) \cap (A \times C) = \{(1, 4), (2, 4), (3, 4)\}$.

(A) First,find the union of sets $B$ and $C$:
$(B \cup C) = \{3, 4\} \cup \{4, 5, 6\} = \{3, 4, 5, 6\}$.
Now,find the Cartesian product $A \times (B \cup C)$:
$A \times (B \cup C) = \{1, 2, 3\} \times \{3, 4, 5, 6\} = \{(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)\}$.

(D) First,we find the Cartesian products $A \times B$ and $A \times C$:
$A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)\}$
$A \times C = \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}$
Now,we find the union $(A \times B) \cup (A \times C)$ by combining all unique ordered pairs from both sets:
$(A \times B) \cup (A \times C) = \{(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)\}$

(N/A) Given $P = \{1, 2\}$.
To find the Cartesian product $P \times P \times P$,we find all possible ordered triples $(a, b, c)$ where $a, b, c \in P$.
$P \times P \times P = \{(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)\}$.

(B) The Cartesian product $R \times R$ is defined as the set $R \times R = \{(x, y) : x, y \in R \}$. This set represents the coordinates of all points in a two-dimensional plane.
The Cartesian product $R \times R \times R$ is defined as the set $R \times R \times R = \{(x, y, z) : x, y, z \in R \}$. This set represents the coordinates of all points in three-dimensional space.

(A) The Cartesian product $A \times B$ is defined as the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$.
Given $A \times B = \{(p, q), (p, r), (m, q), (m, r)\}$.
$A$ is the set of all first elements of the ordered pairs in $A \times B$.
Thus,$A = \{p, m\}$.
$B$ is the set of all second elements of the ordered pairs in $A \times B$.
Thus,$B = \{q, r\}$.

(A) Given that $\left(\frac{x}{3}+1, y-\frac{2}{3}\right)=\left(\frac{5}{3}, \frac{1}{3}\right)$.
Since the ordered pairs are equal,their corresponding elements must be equal.
Equating the first components: $\frac{x}{3}+1=\frac{5}{3}$
$\Rightarrow \frac{x}{3}=\frac{5}{3}-1$
$\Rightarrow \frac{x}{3}=\frac{2}{3}$
$\Rightarrow x=2$
Equating the second components: $y-\frac{2}{3}=\frac{1}{3}$
$\Rightarrow y=\frac{1}{3}+\frac{2}{3}$
$\Rightarrow y=\frac{3}{3}$
$\Rightarrow y=1$
Therefore,the values are $x=2$ and $y=1$.

(C) Given that the number of elements in set $A$ is $n(A) = 3.$
The set $B$ is given as $\{3, 4, 5\},$ so the number of elements in set $B$ is $n(B) = 3.$
The number of elements in the Cartesian product $(A \times B)$ is given by the formula $n(A \times B) = n(A) \times n(B).$
Substituting the values,we get $n(A \times B) = 3 \times 3 = 9.$
Thus,the number of elements in $(A \times B)$ is $9$.

(A) Given $G = \{7, 8\}$ and $H = \{5, 4, 2\}$.
The Cartesian product $P \times Q$ of two non-empty sets $P$ and $Q$ is defined as the set of all ordered pairs $(p, q)$ such that $p \in P$ and $q \in Q$.
Therefore,$G \times H = \{(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)\}$.
Similarly,$H \times G = \{(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)\}$.

(N/A) False.
The Cartesian product $P \times Q$ is defined as the set of all ordered pairs $(p, q)$ such that $p \in P$ and $q \in Q$.
Given $P = \{m, n\}$ and $Q = \{n, m\}$,the Cartesian product is:
$P \times Q = \{(m, n), (m, m), (n, n), (n, m)\}$.

(A) The statement is True.
By the definition of the Cartesian product of two sets $A$ and $B$,$A \times B = \{(x, y) : x \in A \text{ and } y \in B\}$. If $A$ and $B$ are non-empty,there exists at least one element $a \in A$ and $b \in B$,which implies $(a, b) \in A \times B$. Thus,$A \times B$ is a non-empty set.

(B) The statement is True.
Step $1$: Evaluate the intersection $B \cap \varnothing$. Since the intersection of any set with the empty set is the empty set,$B \cap \varnothing = \varnothing$.
Step $2$: Substitute this into the expression: $A \times \{B \cap \varnothing\} = A \times \{\varnothing\}$.
Step $3$: The Cartesian product $A \times \{\varnothing\}$ is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in \{\varnothing\}$. This results in $\{(1, \varnothing), (2, \varnothing)\}$.
Wait,let us re-evaluate the original expression: $A \times \{B \cap \varnothing\}$.
Since $B \cap \varnothing = \varnothing$,the expression becomes $A \times \{\varnothing\}$.
This is $NOT$ equal to $\varnothing$. Therefore,the statement is False.
Corrected statement: If $A = \{1, 2\}$ and $B = \{3, 4\}$,then $A \times \{B \cap \varnothing\} = \{(1, \varnothing), (2, \varnothing)\}$.

(A) It is known that for any non-empty set $A$,the Cartesian product $A \times A \times A$ is defined as the set of all ordered triples $(a, b, c)$ such that $a, b, c \in A$.
Given $A = \{-1, 1\}$.
Therefore,$A \times A \times A = \{(a, b, c) : a, b, c \in \{-1, 1\} \}$.
By taking all possible combinations of elements from $A$ for the three positions,we get:
$A \times A \times A = \{(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)\}$.

(A) Given that $A \times B = \{(a, x), (a, y), (b, x), (b, y)\}$.
We know that the Cartesian product of two non-empty sets $A$ and $B$ is defined as $A \times B = \{(p, q) : p \in A, q \in B\}$.
Therefore,$A$ is the set of all first elements of the ordered pairs in $A \times B$,and $B$ is the set of all second elements.
Thus,$A = \{a, b\}$ and $B = \{x, y\}$.

(N/A) To verify: $A \times (B \cap C) = (A \times B) \cap (A \times C)$.
First,find the intersection of sets $B$ and $C$:
$B \cap C = \{1, 2, 3, 4\} \cap \{5, 6\} = \varnothing$.
Now,calculate the Left Hand Side $(L.H.S.)$:
$L.H.S. = A \times (B \cap C) = \{1, 2\} \times \varnothing = \varnothing$.
Next,calculate the Right Hand Side $(R.H.S.)$:
$A \times B = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)\}$.
$A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\}$.
$R.H.S. = (A \times B) \cap (A \times C) = \varnothing$.
Since $L.H.S. = R.H.S. = \varnothing$,the identity $A \times (B \cap C) = (A \times B) \cap (A \times C)$ is verified.

(N/A) To verify: $A \times C$ is a subset of $B \times D$.
First,calculate the Cartesian product $A \times C$:
$A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\}$.
Next,calculate the Cartesian product $B \times D$:
$B \times D = \{(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)\}$.
We observe that every element of $A \times C$ is present in $B \times D$.
Therefore,$A \times C \subset B \times D$ is verified.

Given $A = \{1, 2\}$ and $B = \{3, 4\}$.
$A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$.
The number of elements in $A \times B$ is $n(A \times B) = 4$.
The number of subsets of a set with $m$ elements is $2^m$.
Therefore,the number of subsets of $A \times B$ is $2^4 = 16$.
The subsets are:
$\varnothing, \{(1, 3)\}, \{(1, 4)\}, \{(2, 3)\}, \{(2, 4)\}, \{(1, 3), (1, 4)\}, \{(1, 3), (2, 3)\}, \{(1, 3), (2, 4)\}, \{(1, 4), (2, 3)\}, \{(1, 4), (2, 4)\}, \{(2, 3), (2, 4)\}, \{(1, 3), (1, 4), (2, 3)\}, \{(1, 3), (1, 4), (2, 4)\}, \{(1, 3), (2, 3), (2, 4)\}, \{(1, 4), (2, 3), (2, 4)\}, \{(1, 3), (1, 4), (2, 3), (2, 4)\}$.

(A) Given that $n(A) = 3$ and $n(B) = 2$.
Since $(x, 1), (y, 2), (z, 1) \in A \times B$,the first elements of these ordered pairs must belong to set $A$ and the second elements must belong to set $B$.
Thus,$A = \{x, y, z\}$ and $B = \{1, 2\}$.
Since $n(A) = 3$ and $n(B) = 2$,these sets satisfy the given conditions.
Therefore,$A = \{x, y, z\}$ and $B = \{1, 2\}$.

(A) We know that if $n(A) = p,$ then $n(A \times A) = p^2.$
Given $n(A \times A) = 9,$ so $p^2 = 9,$ which implies $n(A) = 3.$
The elements of $A \times A$ are of the form $(a, b)$ where $a, b \in A.$
Since $(-1, 0) \in A \times A,$ we have $-1 \in A$ and $0 \in A.$
Since $(0, 1) \in A \times A,$ we have $0 \in A$ and $1 \in A.$
Thus,the set $A = \{-1, 0, 1\}.$
The set $A \times A$ consists of $3 \times 3 = 9$ elements:
$A \times A = \{(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)\}.$
The given elements are $(-1, 0)$ and $(0, 1).$
The remaining elements are $(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1).$

(C) Given $n(A)=4$ and $n(B)=2$.
The number of elements in the Cartesian product $A \times B$ is $n(A \times B) = n(A) \times n(B) = 4 \times 2 = 8$.
We need to find the number of subsets of $A \times B$ having at least $3$ elements.
The total number of subsets of a set with $8$ elements is $2^8 = 256$.
The number of subsets with $0, 1,$ or $2$ elements are given by $\binom{8}{0} + \binom{8}{1} + \binom{8}{2}$.
$\binom{8}{0} = 1$,$\binom{8}{1} = 8$,and $\binom{8}{2} = \frac{8 \times 7}{2} = 28$.
Sum of these subsets $= 1 + 8 + 28 = 37$.
The number of subsets with at least $3$ elements $= 2^8 - (\binom{8}{0} + \binom{8}{1} + \binom{8}{2}) = 256 - 37 = 219$.

(C) $A \cap B = \{4\}$
$A \cup B = \{2, 3, 4, 5\}$
Therefore,the Cartesian product is:
$(A \cap B) \times (A \cup B) = \{4\} \times \{2, 3, 4, 5\}$
$= \{(4, 2), (4, 3), (4, 4), (4, 5)\}$

(A) Given,$n(A)=p, n(B)=q$ and $n(A \times B)=7$.
Since,$n(A \times B)=n(A) \times n(B)$,we have $p \times q = 7$.
Since $p$ and $q$ are the number of elements in sets,they must be positive integers. The factors of $7$ are $1$ and $7$.
Thus,the possible values for $(p, q)$ are $(7, 1)$ or $(1, 7)$.
In both cases,$p^2+q^2 = 7^2 + 1^2 = 49 + 1 = 50$.