Question

Given $A=\left\{ 2,3,5 \right\}$ and $B=\left\{ 0,1 \right\}$. Find the number of different ordered pairs in which the first entry is an element of A and the second is an element of B.

Answer 1

Hint: Now take the pair as $\left( a,b \right)$ where $a\in A,b\in B$. From the elements of B, take zero and put it in the pair to make the ordered pair set $\left( A,B \right)$. We do the similar thing for $1\in B$. Thus, we find the number of pairs that can be formed. We take the number of elements $\left( a,b \right)$ in $\left( A,B \right)$ as $n\left( A,B \right)$ and that number will be the solution of the problem.

Complete step-by-step answer:
We have been given $A=\left\{ 2,3,5 \right\}$ and $B=\left\{ 0,1 \right\}$.
Now, we need to find the number of different ordered pairs for which \[{{1}^{st}}\] entry is an element from A and the \[{{2}^{nd}}\] is an element from B. We need to find a pair, and no repetition is allowed. We can represent the set as $\left( A,B \right)$ with elements being of the form $\left( a,b \right)$.
The form of the set will be $\left( a,b \right)$ where $a\in A,b\in B$.
Let us consider $A=\left\{ 2,3,5 \right\}$ and $B=\left\{ 0,1 \right\}$.
Now let us put B = 0 in the pair $\left( A,B \right)$. Thus, we can form the pairs $\left( 2,0 \right),\left( 3,0 \right),\left( 5,0 \right)$.
We have taken the preceding points like 2, 3, 5 in the dual from set A.
\[\therefore \] By putting B as zero, we get three pairs of elements. Hence with set A there are 3 possible ways for each element in B. Now we are told that there is no repetition.
Thus, with B as 1, we can make 3 pairs as we did for B = 0. Hence the 3 possible ways are $\left( 2,1 \right),\left( 3,1 \right),\left( 5,1 \right)$.
Thus, the total number of possible pairs 6 possible ways.
The 6 possible elements of the set $\left( A,B \right)$ are $\left( 2,0 \right),\left( 3,0 \right),\left( 5,0 \right),\left( 2,1 \right),\left( 3,1 \right),\left( 5,1 \right)$.
Hence, we got the number of different ordered pairs $n\left( A,B \right)=6$.

Note: We can also solve it in this method,
As there are 3 elements in A, $n\left( A \right)=3$. There are 2 elements in B, $n\left( B \right)=2$.
Now take the pair as $\left( a,b \right)$ where $a\in A,b\in B$. We need to find the number of elements in $n\left( A,B \right)$.
\[\therefore \] Total possible pairs = \[n\left( A,B \right)=n\left( A \right)\times n\left( B \right)=3\times 2=6\] pairs.