Question

Let $n\left( A \right)=n$ . Then the number of all relation on A is:
(a). ${{2}^{n}}$
(b). ${{2}^{n!}}$
(c). ${{2}^{{{n}^{2}}}}$
(d). None

Answer 1

Hint: We are going to solve this question by putting different values of n and then we will find the number of subset possible and that will be equal to all relations and then we will try to find the formula for n.

Complete step-by-step solution -

For n(A) = 1,
Let A = $\{1\}$
Then the total subset will be $\left\{ 1,\phi \right\}$
Hence 2 elements for n = 1.
Now for n(A) = 2,
Let A = { 1, 2 }
The total subset will be $\left\{ 1,2,\left( 1,2 \right),\phi \right\}$
Hence 4 elements for n = 2.
Now for n(A) = 3,
Let A = { 1, 2, 3 }
The total subset will be $\left\{ 1,2,3,\left( 2,3 \right),\left( 1,3 \right),\left( 1,2 \right),\left( 1,2,3 \right),\phi \right\}$
Hence 8 elements for n = 3.
So, from this we can see that ${{2}^{n}}$ is the correct formula.
We also know that if a set has n elements then the number of subsets of A are ${{2}^{n}}$ ,
Therefore, for $A\times A$ there can be as many relations as the number of subsets of $A\times A$.
Hence, the correct answer is option (a).

Note: In this question we have derived the formula for total number of subset and in that derivation we have considered an empty element $\phi $ one can also directly use the formula that the number of subset of set having n elements is ${{2}^{n}}$, most of the students miss the empty element $\phi $ and then make mistakes while solving, so these point should be kept in mind.