Let A = {1, 2, 3, ldots, 7} and let P(A) deno | vedclass

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(D) The set $A$ has $7$ elements,so the power set $P(A)$ has $2^7 = 128$ elements.For each element $a \in A$,we must choose a subset $f(a) \subseteq A

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$44$

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(D) The set $A$ has $7$ elements,so the power set $P(A)$ has $2^7 = 128$ elements.For each element $a \in A$,we must choose a subset $f(a) \subseteq A$ such that $a \in f(a)$.The number of subsets of $A$ that contain a specific element $a$ is $2^{7-1} = 2^6 = 64$.Since there are $7$ elements in $A$,and for each element $a$,there are $2^6$ choices for $f(a)$,the total number of such functions is $(2^6) 	imes (2^6) 	imes (2^6) 	imes (2^6) 	imes (2^6) 	imes (2^6) 	imes (2^6) = (2^6)^7 = 2^{42}$.We are given that the number of functions is $m^n$ where $m$ is the least value.Since $2^{42} = (2^2)^{21} = 4^{21}$ and $2^{42} = (2^1)^{42} = 2^{42}$,the least base $m$ is $2$ with $n = 42$.Thus,$m + n = 2 + 42 = 44$.

Let $A = \{1, 2, 3, \ldots, 7\}$ and let $P(A)$ denote the power set of $A$. If the number of functions $f: A \rightarrow P(A)$ such that $a \in f(a)$ for all $a \in A$ is $m^n$,where $m, n \in N$ and $m$ is the least possible value,then $m + n$ is equal to . . . . . . .

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