Question

Let A and B two sets such that $n\left[ P\left( A \right) \right]=n\left[ P\left( B \right) \right]$ If (a,2), (b,3), (c,3) are in ... Find A and B where a, b, c are distinct. Find $A\times B$ ?

Answer 1

First at all, we should know the expression used in the above question, i.e. $n\left[ P\left( A \right) \right]\text{ and }n\left[ P\left( B \right) \right]$. It is basically related to power sets and its properties. Once we know the formula for these expressions, then we will easily identify the clue or approach for getting a solution. If we find out the number of elements in set A and set B, then by using the data given for $A\times B$ we can find set A and set B and further find $A\times B$ fully. In the question, the data given for $A\times B$ is partly why it is asking for $A\times B$ at the end of the question.

Complete step-by-step answer:
Now, let us see the given data:

& n\left[ P\left( A \right) \right]=8 \\\ & n\left[ P\left( B \right) \right]=4 \\\ \end{aligned}$$ (a,2), (b,3), (c,3) are in $A\times B$ We know, if A is any set then the set of all subsets of A is called the power set of A and is denoted by P(A). $n\left[ P\left( A \right) \right]$ is the number of elements in the power set P(A). We know the formula for this, i.e. $$\Rightarrow n\left[ P\left( A \right) \right]={{2}^{x}}$$ Where, x is the number of elements in set (A) Hence, from given data, $$\begin{aligned} & \Rightarrow n\left[ P\left( A \right) \right]={{2}^{x}}=8 \\\ & \Rightarrow {{2}^{x}}={{2}^{3}} \\\ \end{aligned}$$ By comparing we get x = 3. Hence, the number of elements in set A would be 3. Similarly, $$\begin{aligned} & \Rightarrow n\left[ P\left( B \right) \right]={{2}^{y}}=4 \\\ & \Rightarrow {{2}^{y}}={{2}^{2}} \\\ \end{aligned}$$ By comparing we have y = 2. Hence, the number of elements in set B would be 2. Let, $$\text{Set }A=\left\\{ {{x}_{1}},{{x}_{2}},{{x}_{3}} \right\\}\text{ and Set }B=\left\\{ {{y}_{1}},{{y}_{2}} \right\\}$$ Hence, $$A\times B=\left\\{ \left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{1}},{{y}_{2}} \right),\left( {{x}_{2}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{1}} \right),\left( {{x}_{3}},{{y}_{2}} \right) \right\\}\cdots \cdots \left( i \right)$$ Given data, $$\begin{matrix} A\times B=\left\\{ \cdots \left( a,2 \right)\cdots \left( b,3 \right)\cdots \left( c,3 \right) \right\\}\cdots \cdots \left( ii \right) \\\ \downarrow \\\ \text{Pairs are randomly placed} \\\ \end{matrix}$$ Don’t focus on which location they are located. This is just the sample picture. By comparing (ii) with (i), we will get: $$\begin{aligned} & {{x}_{1}}=a \\\ & {{x}_{2}}=b \\\ & {{x}_{3}}=c \\\ & {{y}_{1}}=2 \\\ & {{y}_{2}}=3 \\\ \end{aligned}$$ Where a, b and c are distinct. Hence, we have $$\text{Set A}=\left\\{ a,b,c \right\\}\text{ and Set B}=\left\\{ 2,3 \right\\}$$ Therefore, $$\text{Set }A\times B=\left\\{ \left( a,2 \right),\left( a,3 \right),\left( b,2 \right),\left( b,3 \right),\left( c,2 \right),\left( c,3 \right) \right\\}$$ **Note:** Look in above solution, we have taken $\left( {{x}_{1}}=a \right)$ but it can also be b or c. Because in question, there is no discussion about whether (a,2), (b,3), (c,3) are present in ordered way or not in set $A\times B$ One important thing whose students should be aware in case of $A\times B$ i.e. Cartesian product $$C=A\times B$$ Let, A = {a,b} and B = {0,1} $$C=A\times B=\left\\{ \left( a,0 \right),\left( a,1 \right),\left( b,0 \right),\left( b,1 \right) \right\\}$$ We can’t change $\left( a,0 \right)\to \left( 0,a \right)$ but we can change like $\left\\{ \left( a,0 \right),\left( a,1 \right),\left( b,0 \right),\left( b,1 \right) \right\\}$