Key Concepts and Formulas

• Power Set: The power set of a set A, denoted by P(A), is the set of all subsets of A. If $|A| = k$, then $|P(A)| = 2^k$.
• Functions: A function $f: A \rightarrow B$ assigns to each element in set A exactly one element in set B.
• Multiplication Principle: If there are $n_1$ ways to do one thing, $n_2$ ways to do another, ..., and $n_k$ ways to do a $k$-th thing, then the total number of ways to do all $k$ things in sequence is $n_1 \times n_2 \times \ldots \times n_k$.
• Counting Subsets Containing a Specific Element: The number of subsets of a set with $k$ elements that contain a specific element is $2^{k-1}$.

Step-by-Step Solution

Step 1: Understand the Domain, Codomain, and the Function's Nature The domain of the function is the set $\mathrm{A} = \{1, 2, 3, 4, 5, 6, 7\}$. The number of elements in the domain is $|\mathrm{A}| = 7$. The codomain of the function is the power set of $\mathrm{A}$, denoted by $\mathrm{P}(\mathrm{A})$. This means that for any element $\mathrm{a} \in \mathrm{A}$, its image $f(\mathrm{a})$ must be a subset of $\mathrm{A}$. The total number of elements in the codomain is $|\mathrm{P}(\mathrm{A})| = 2^{|\mathrm{A}|} = 2^7 = 128$.

Step 2: Analyze the Constraint on the Function Mapping The problem imposes a specific condition: $\mathrm{a} \in f(\mathrm{a})$ for all $\mathrm{a} \in \mathrm{A}$. This means that for every element in the domain, its image (which is a subset of $\mathrm{A}$) must contain that element itself.

Step 3: Determine the Number of Possible Images for Each Element in the Domain Let's consider an arbitrary element $\mathrm{a} \in \mathrm{A}$. We need to find how many subsets of $\mathrm{A}$ satisfy the condition $\mathrm{a} \in f(\mathrm{a})$. To form such a subset $f(\mathrm{a})$, the element $\mathrm{a}$ must be included. The remaining elements of the subset can be any combination of elements from the set $\mathrm{A} \setminus \{\mathrm{a}\}$. The set $\mathrm{A} \setminus \{\mathrm{a}\}$ has $|\mathrm{A}| - 1 = 7 - 1 = 6$ elements. The number of ways to choose any subset from these 6 elements is $2^6$. For each of these $2^6$ choices of subsets from $\mathrm{A} \setminus \{\mathrm{a}\}$, when we add the element $\mathrm{a}$ to it, we get a unique subset of $\mathrm{A}$ that contains $\mathrm{a}$. Therefore, for each element $\mathrm{a} \in \mathrm{A}$, there are $2^6$ possible choices for $f(\mathrm{a})$.

Step 4: Calculate the Total Number of Such Functions Since the choice of the image $f(\mathrm{a})$ for each element $\mathrm{a} \in \mathrm{A}$ is independent of the choices for other elements, we can use the multiplication principle. There are 7 elements in set A: 1, 2, 3, 4, 5, 6, 7. For each of these 7 elements, there are $2^6$ possible images. Total number of functions = (Number of choices for $f(1)$) $\times$ (Number of choices for $f(2)$) $\times \ldots \times$ (Number of choices for $f(7)$) Total number of functions = $2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6 = (2^6)^7$.

Using the exponent rule $(a^b)^c = a^{b \times c}$, we get: Total number of functions = $2^{6 \times 7} = 2^{42}$.

Step 5: Determine $m+n$ We are given that the number of functions is $\mathrm{m}^{\mathrm{n}}$, where $\mathrm{m}, \mathrm{n} \in \mathrm{N}$ and $\mathrm{m}$ is the least possible. We found the number of functions to be $2^{42}$. To satisfy the condition that $\mathrm{m}$ is the least natural number, we set $\mathrm{m} = 2$ and $\mathrm{n} = 42$. Here, $m=2$ is the smallest prime base, and thus the least possible base for this expression. We need to find $\mathrm{m}+\mathrm{n}$. $\mathrm{m}+\mathrm{n} = 2 + 42 = 44$.

Common Mistakes & Tips

• Confusing elements and subsets: Ensure you differentiate between an element $\mathrm{a}$ from set $\mathrm{A}$ and its image $f(\mathrm{a})$, which is a subset of $\mathrm{A}$ (an element of $\mathrm{P}(\mathrm{A})$).
• Incorrectly counting subsets: When counting subsets that must contain a specific element, remember to consider the remaining elements from the set excluding that specific element. The number of such subsets is $2^{k-1}$ for a set of size $k$.
• Base $m$ being least: When expressing a number in the form $m^n$ with the least $m$, use the smallest prime factor as the base if possible. For $2^{42}$, the least base is $2$.

Summary The problem required us to count the number of functions $f: \mathrm{A} \rightarrow \mathrm{P}(\mathrm{A})$ such that $\mathrm{a} \in f(\mathrm{a})$ for all $\mathrm{a} \in \mathrm{A}$. We determined that for each of the 7 elements in set A, there are $2^{7-1} = 2^6$ possible subsets it can map to, satisfying the condition. By applying the multiplication principle, the total number of such functions is $(2^6)^7 = 2^{42}$. Given this is in the form $m^n$ with the least $m$, we have $m=2$ and $n=42$. Therefore, $m+n = 2+42 = 44$.

The final answer is $\boxed{44}$.