Question

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 $a \in f(a), \, \forall a \in A$ is $m^n$, $m$ and $n \in \mathbb{N}$ and $m$ is least, then $m + n$ is equal to \\_\\_\\_\\_\\_\\_\\_\\_\\_.

Answer 1

Solution: Each element $a$ in $A$ must be included in its corresponding subset in $P(A)$, so we only consider subsets of $A$ that contain $a$. For each element, there are $2^6$ possible subsets of $A$ that include $a$ (since we can select or omit any of the remaining 6 elements).

Thus, for each $a \in A$, there are $2^6$ choices, and since there are 7 elements in $A$:

Total number of functions = $(2^6)^7 = 2^{42}$

Since we need $m^n = 2^{42}$ with $m$ and $n$ as small as possible:

$m = 2$, $n = 42$

Therefore, $m + n = 2 + 42 = 44$.