If \[{\text{A = }}\left\{ {{\text{a,b,c,d}}} \right\}\], \[{\text{B = }}\left\{ {{\text{p,q,r,s}}} \right\}\]then which of the following are relations from A to B?
 A) \[R1 = \left\{ {\left( {a,p} \right),\left( {b,q} \right),\left( {c,s} \right)} \right\}\]
B) \[R2 = \left\{ {\left( {q,b} \right),\left( {c,s} \right),\left( {d,r} \right)} \right\}\]
C) \[R3 = \left\{ {\left( {a,p} \right),\left( {a,q} \right),\left( {d,p} \right),\left( {c,r} \right),\left( {d,r} \right)} \right\}\]
D) \[R4 = \left\{ {\left( {a,p} \right),\left( {q,a} \right),\left( {b,s} \right),\left( {s,b} \right)} \right\}\]

Hint: Here we have to let a relation R from A to B such that \[{\text{R}} \subseteq {\text{A}} \times {\text{B}}\] and the domain of
\[{\text{R = \{ a:a}} \in {\text{A (a,b)}} \in {{R\;for some\;b}} \in {\text{B\} }}\]and the range of \[{\text{R = \{ \;b:b}} \in {\text{B,(a,b)}} \in {{R\;for some\;a}} \in {\text{A\} }}\]
And then calculate \[{\text{A}} \times {\text{B}}\]

Complete step by step solution:
We are given two sets \[{\text{A = }}\left\{ {{\text{a,b,c,d}}} \right\}\] and \[{\text{B = }}\left\{ {{\text{p,q,r,s}}} \right\}\] and we have to make a relation from A to B.
Let the relation from A to B be R such that \[{\text{R}} \subseteq {\text{A}} \times {\text{B}}\] and the domain of
\[{\text{R = \{ a:a}} \in {\text{A (a,b)}} \in {{R\;for some\;b}} \in {\text{B\} }}\]and the range of \[{\text{R = \{ \;b:b}} \in {\text{B,(a,b)}} \in {{R\;for some\;a}} \in {\text{A\} }}\]
Then ,
First calculating \[{\text{A}} \times {\text{B}}\] we get:
\[{\text{A}} \times {\text{B}} = \left\{ {\left( {{\text{a,p}}} \right){\text{,}}\left( {a,q} \right),\left( {a,r} \right),\left( {a,s} \right),\left( {b,p} \right),\left( {{\text{b,q}}} \right){\text{,}}\left( {b,r} \right){\text{,}}\left( {b,s} \right){\text{,}}\left( {c,p} \right){\text{,}}\left( {c,q} \right){\text{,}}\left( {{\text{c,r}}} \right){\text{,}}\left( {c,s} \right){\text{,}}\left( {d,p} \right){\text{,}}\left( {d,q} \right){\text{,}}\left( {d,r} \right){\text{,}}\left( {{\text{d,s}}} \right)} \right\}\]
Since R should be subset of \[{\text{A}} \times {\text{B}}\]
\[{\text{R}} \subseteq {\text{A}} \times {\text{B}}\]
Therefore checking for option A we get:-
\[R1 = \left\{ {\left( {a,p} \right),\left( {b,q} \right),\left( {c,s} \right)} \right\}\]
Since,
\[{\text{R1}} \subseteq {\text{A}} \times {\text{B}}\]
Therefore, option A is correct.
Now checking for option B we get:-
\[R2 = \left\{ {\left( {q,b} \right),\left( {c,s} \right),\left( {d,r} \right)} \right\}\]
Since,
\[{\text{R2}} \not\subset {\text{A}} \times {\text{B}}\]
Therefore option B is incorrect.
Checking for option C we get:-
\[R3 = \left\{ {\left( {a,p} \right),\left( {a,q} \right),\left( {d,p} \right),\left( {c,r} \right),\left( {d,r} \right)} \right\}\]
Since,
\[{\text{R3}} \subseteq {\text{A}} \times {\text{B}}\]
Therefore it is a correct option.
Now checking for option D we get:-
\[R4 = \left\{ {\left( {a,p} \right),\left( {q,a} \right),\left( {b,s} \right),\left( {s,b} \right)} \right\}\]
Since,
\[{\text{R4}} \not\subset {\text{A}} \times {\text{B}}\]
Therefore it is incorrect option.

Hence, option A and option C are correct options.

Note:
The relation R should be a subset of \[{\text{A}} \times {\text{B}}\] to be a relation from A to B in which the first element of the ordered pair should be from set A and the second element should be from set B.