$$\eqalign{ & \left\{ {\left( {a,\,1} \right),\,\left( {a,\,3} \right)} \right\} \subseteq A \times B\,\,\therefore \,{\text{This is a relation}} \cr & \left\{ {\left( {b,\,1} \right),\,\left( {c,\,2} \right),\,\left( {d,\,1} \right)} \right\} \subseteq A \times B\,\,\therefore \,{\text{This is a relation}} \cr & \left( {3,\,b} \right)\, \notin \,A \times B \cr & \therefore \,\left\{ {\left( {a,\,2} \right),\,\left( {b,\,3} \right),\,\left( {3,\,b} \right)} \right\}{\text{ is not a relation from }}A{\text{ to }}B \cr & \left\{ {\left( {a,\,1} \right),\,\left( {b,\,2} \right),\,\left( {c,\,3} \right)} \right\}\,\,{\text{is a relation from }}A{\text{ to }}B \cr} $$

Let $$x \in A\,\,{\text{and }}x \in B$$
$$\eqalign{ & \Leftrightarrow \,x \in A \cup B \cr & \Leftrightarrow \,\,x \in A \cup C\,\,\,\left( {\because A \cup B = A \cup C} \right) \cr & \Leftrightarrow x \in C \cr & \therefore \,\,B = C. \cr} $$
Let $$x \in A\,\,{\text{and }}x \in B$$
$$\eqalign{ & \Leftrightarrow \,x \in A \cap B \cr & \Leftrightarrow \,\,x \in A \cap C\,\,\,\left( {\because A \cap B = A \cap C} \right) \cr & \Leftrightarrow x \in C \cr & \therefore \,\,B = C. \cr} $$

The given function $$f\left( x \right) = {}^{7 - x}{P_{x - 3}}$$    would be defined, if
$$\eqalign{ & ({\text{i}})\,7 - x > 0\, \Rightarrow x < 7 \cr & ({\text{ii}})\,x - 3 \geqslant 0\, \Rightarrow x \geqslant 3 \cr & ({\text{iii}})\,\left( {x - 3} \right) \leqslant \left( {7 - x} \right) \Rightarrow 2x \leqslant 10\, \Rightarrow x \leqslant 5\, \Rightarrow x = 3,\,4,\,5, \cr & {\text{Hence, range of }}\,f\left( x \right) = \left\{ {{}^4{P_0},\,{}^3{P_1},\,{}^2{P_2}} \right\} \cr & {\text{Range of }}\,f\left( x \right) = \left\{ {1,\,3,\,2} \right\} \cr} $$

Let a Venn diagram be drawing taking three intersecting sets $$A,\,B$$  and $$C$$ under the universal set $$U.$$ This makes 8 regions $$a$$ to $$h$$ as shown.

$$A$$ has regions $$a,\,b,\,d,\,e$$
$$B$$ has regions $$b,\,c,\,e,\,f$$
$$C$$ has regions $$d,\,e,\,f,\,g$$
$$C'$$ has regions $$a,\,b,\,c,\,h$$
$$B'$$ has regions $$a,\,d,\,g,\,h$$
$$\eqalign{ & {\bf{Statement (A)}}\,{\bf{:}}\,A \cup \left( {B - C} \right) = A \cap \left( {B \cap C} \right) \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \equiv \left( {a,\,b,\,e,\,d} \right) \cup b,\,c \equiv a,\,b,\,c,\,d,\,e \cr & {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \equiv a,\,b,\,d,\,e \cap e,\,f \equiv e \cr & {\text{So, statement (A) is not correct}}{\text{. }} \cr & {\bf{Statement (B)}}\,{\bf{:}}\,A - \left( {B \cup C} \right) = \left( {A \cap B'} \right) \cap C' \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \equiv \left( {a,\,b,\,d,\,e} \right) - \left( {b,\,c,\,d,\,e,\,f,\,g} \right) \equiv a \cr & {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \equiv \left( {a,\,b,\,d,\,e \cap a,\,d,\,g,\,h} \right) \cap \left( {a,\,b,\,c,\,h} \right) \equiv a \cr & {\text{So, statement (B) is correct}}{\text{.}} \cr & {\text{Correct statement is : }}A - \left( {B \cup C} \right) = \left( {A \cap B'} \right) \cap C' \cr} $$

$$\eqalign{ & x\, \in \,R \Rightarrow x - x + \sqrt 5 = \sqrt 5 {\text{ is an irrational number}}{\text{.}} \cr & \therefore \,\left( {x,\,x} \right) \in \,R \cr & \therefore \,R{\text{ is reflexive}}{\text{.}} \cr & \left( {\sqrt 5 ,\,1} \right) \in \,R{\text{ because}} \cr & \sqrt 5 - 1 + \sqrt 5 = 2\sqrt 5 - 1{\text{ which is an irrational number}}{\text{.}} \cr & \therefore \,\left( {1,\,\sqrt 5 } \right) \notin \,R \cr & \therefore \,R{\text{ is not symmetric}}{\text{.}} \cr & {\text{We have, }}\left( {\sqrt 5 ,\,1} \right),\,\left( {1,\,2\sqrt 5 } \right) \in \,R{\text{ because}} \cr & \sqrt 5 - 1 + \sqrt 5 = 2\sqrt 5 - 1{\text{ if }}1 - 2\sqrt 5 + \sqrt 5 = 1 - \sqrt 5 \cr & {\text{are irrational number}}{\text{.}} \cr & {\text{Also, }}\left( {\sqrt 5 ,\,2\sqrt 5 } \right) \in \,R{\text{ and}} \cr & \sqrt 5 - 2\sqrt 5 + \sqrt 5 = 0\,{\text{which is not an irrational number}}{\text{.}} \cr & \therefore \,\left( {\sqrt 5 ,\,2\sqrt 5 } \right) \notin \,R \cr & \therefore \,R{\text{ is not transitive}}{\text{.}} \cr} $$

Clearly $$(x, x)$$  $$ \in R\,\forall x \in W.$$   So $$R$$ is relexive.
Let $$(x, y)$$  $$ \in R,$$  then $$(y, x)$$  $$ \in R$$  as $$x$$ and $$y$$ have at least one letter in common. So, $$R$$ is symmetric.
But $$R$$ is not transitive for example
Let $$x$$ = INDIA, $$y$$ = BOMBAY and $$z$$ = JOKER
then $$(x, y)$$  $$ \in R$$  ($$A$$ is common) and $$(y, z)$$  $$ \in R$$  ($$O$$ is common) but $$(x, z)$$  $$ \notin R.$$  (as no letter is common)

Total number of persons $$ = a + b + c + n = 100$$

Do not prefer fish $$b + n = 50$$
$$60$$  prefer chicken hence $$b + c = 60$$
Do not like fish and chicken is $$n = 10$$
On solving these equations we will get $$a = 30,\,b = 40,\,c = 20$$
The number of persons who prefer both fish and chicken is $$ = c = 20$$

$$\eqalign{ & R = \left\{ {\left( {x,\,y} \right):x,\,y\, \in \,N{\text{ and }}{x^2} - 4xy + 3{y^2} = 0} \right\}, \cr & {\text{Now, }}{x^2} - 4xy + 3{y^2} = 0\,\,\, \Rightarrow \left( {x - y} \right)\left( {x - 3y} \right) = 0 \cr & \therefore \,x = y{\text{ or }}x = 3y \cr & \therefore \,R = \left\{ {\left( {1,\,1} \right),\,\left( {3,\,1} \right),\,\left( {2,\,2} \right),\,\left( {6,\,2} \right),\,\left( {3,\,3} \right),\,\left( {9,\,3} \right),.....} \right\} \cr} $$
Since $$\left( {1,\,1} \right),\,\left( {2,\,2} \right),\,\left( {3,\,3} \right),.....$$       are present in the relation, therefore $$R$$ is reflexive.
Since $$\left( {3,\,1} \right)$$  is an element of $$R$$ but $$\left( {1,\,3} \right)$$  is not the element of $$R$$ is not symmetric.
$$\eqalign{ & {\text{Here }}\left( {3,\,1} \right)\, \in \,R{\text{ and }}\left( {1,\,1} \right)\, \in \,R\, \Rightarrow \left( {3,\,1} \right)\, \in \,R \cr & \left( {6,\,2} \right)\, \in \,R{\text{ and }}\left( {2,\,2} \right)\, \in \,R\, \Rightarrow \left( {6,\,2} \right)\, \in \,R \cr & {\text{For all such }}\left( {a,\,b} \right)\, \in \,R{\text{ and }}\left( {b,\,c} \right)\, \in \,R \Rightarrow \left( {a,\,c} \right)\, \in \,R \cr & {\text{Hence }}R{\text{ is transitive}}{\text{.}} \cr} $$

Number of definitions $$=$$ Number of mapping from $$A$$ to $$B = 3 \times 3 \times 3 = 27.$$

$$\eqalign{ & {}^{24 - x}{C_{3x - 1}}{\text{ is defined if,}} \cr & 24 - x > 0,\,3x - 1 \geqslant 0{\text{ and }}24 - x \geqslant 3x - 1 \cr & \Rightarrow x < 24,\,x \geqslant \frac{1}{3}{\text{ and }}x \leqslant \frac{{25}}{4} \cr & \Rightarrow \frac{1}{3} \leqslant x \leqslant \frac{{25}}{4} \cr & {}^{40 - 6x}{C_{8x - 10}}{\text{ is defined if,}} \cr & 40 - 6x > 0,\,8x - 10 \geqslant 0{\text{ and }}40 - 6x \geqslant 8x - 10 \cr & \Rightarrow x < \frac{{20}}{3},\,x \geqslant \frac{5}{4}{\text{ and }}x \leqslant \frac{{25}}{7} \cr & \Rightarrow \frac{5}{4} \leqslant x \leqslant \frac{{25}}{7} \cr & {\text{From (1) and (2), we get }}\frac{5}{4} \leqslant x \leqslant \frac{{25}}{7} \cr & {\text{But }}24 - x\, \in \,N, \cr & \therefore \,\,x{\text{ must be an integer, }}x = 2,\,3 \cr & {\text{Hence domain }}\left( f \right) = \left\{ {2,\,3} \right\} \cr} $$