Let A = 1, 2, 3, . 7 and let P ( A ) denote the power set of A . If...

Given,

$A = { 1, 2, 3, \ldots . 7 }$

And $f: A \to P ( A )$ and $a \in A$

Also $a \in f ( a )$, that means $‘ a ’$ will be the element of subset as $f: A \to P ( A )$,

Now, total options for $1$ will be ${2}^{6}$. (Because ${2}^{6}$ subsets contains elements$1$)

Similarly, for every other element we get, total  ${2}^{6} \times {2}^{6} \times {2}^{6} \times {2}^{6} \times {2}^{6} \times {2}^{6} \times {2}^{6} = {2}^{42}$ subsets.

Now, comparing with ${m}^{n}$ we get, $m = 2 \text{ and } n = 42$

Hence, $m + n = 2 + 42 = 44$