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Let A be a finite set of size n. The number of elements in the power set of $A\times A$ is:

1. $2^{2^n}$
2. $2^{n^2}$
3. $\left(2^n\right)^2$
4. $\left(2^2\right)^n$
5. None of the above

1 comment

edited

Asking for number of elements in $A \times A$ is same as asking for number of relations possible over set $A.$

Every relation on set $A$ is subset of $A \times A.$ Every subset of $A \times A$ is a relation over set $A$.

Power set of $A \times A$ contains subsets of $A \times A$. So, Every element in power set of $A \times A$ is a relation on set $A.$

Hence, the cardinality of $P(A \times A)$ is same as number of relations possible on set $A.$

If $|A| = n,$ then $|P(A)| = 2^n, |A \times A| = n^2$, So, $|P(A \times A)| = 2^{n^2} = 2^{(n^2)}$

Detailed Video Solution

Cardinality of $A\times A = n^2$
Cardinality of power  set of $A\times A = 2^{n^2}$

Correct Answer: $B$

Say set is A= {1,2,3}=n

Subset of set A ={phi, {1} , {2} , {3} , {1,2} , {2,3}, {1,3} , {1,2,3}}=2n

Now, $A\times A=\left \{ 1,2,3 \right \}\times \left \{ 1,2,3 \right \}$

$=\left \{ \left \{ 1,1 \right \},\left \{ 1,2 \right \} \left \{ 1,3 \right \}\left \{ 2,1 \right \}\left \{ 2,2 \right \}\left \{ 2,3 \right \}\left \{ 3,1 \right \}\left \{ 3,2 \right \}\left \{ 3,3 \right \}\right \}$

So, number of subsets will be$2^{n^{2}}$

This is same as the total number of relations on Set A

$|A| = n$  ,  $|p(A)| = 2^n$

$| A×A| =n^2$

$| p(A×A)|$ =$2^{n^{2}}$

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