Key Concepts and Formulas

• Number of Subsets: A set with $k$ elements has $2^k$ subsets.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a Cartesian plane is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• Properties of Exponents and Factorization: Understanding how to manipulate exponential equations and utilize unique prime factorization is essential.

Step-by-Step Solution

Step 1: Formulate the Equation from the Given Information Let $A$ be a set with $m$ elements and $B$ be a set with $n$ elements. The total number of subsets of set $A$ is $2^m$. The total number of subsets of set $B$ is $2^n$. The problem states that the total number of subsets of set $A$ is 56 more than the total number of subsets of set $B$. This can be written as an equation: $2^m = 2^n + 56$ Rearranging this equation, we get: $2^m - 2^n = 56$

Step 2: Solve the Exponential Equation for $m$ and $n$ Since $2^m - 2^n = 56 > 0$, it implies $2^m > 2^n$, which means $m > n$. We can factor out the common term $2^n$ from the left side of the equation: $2^n (2^{m-n} - 1) = 56$ Now, we need to express 56 as a product of a power of 2 and an odd number. The term $2^n$ is a power of 2, and $(2^{m-n} - 1)$ must be an odd number (since $2^{m-n}$ is even for $m-n \ge 1$, subtracting 1 results in an odd number). Let's find the prime factorization of 56: $56 = 8 \times 7 = 2^3 \times 7$ So, the equation becomes: $2^n (2^{m-n} - 1) = 2^3 \times 7$ By comparing the powers of 2 and the odd factors on both sides of the equation, and using the uniqueness of prime factorization, we can equate them: Equating the powers of 2: $2^n = 2^3$ $n = 3$ Equating the odd factors: $2^{m-n} - 1 = 7$ Now, we solve for $m-n$: $2^{m-n} = 7 + 1$ $2^{m-n} = 8$ Since $8 = 2^3$, we have: $2^{m-n} = 2^3$ $m-n = 3$ We have $n=3$ and $m-n=3$. Substituting $n=3$ into the second equation: $m - 3 = 3$ $m = 6$ Thus, we have found $m=6$ and $n=3$.

Step 3: Determine the Coordinates of Point P The point $P$ is given by $(m, n)$. Substituting the values we found: $P(6, 3)$

Step 4: Calculate the Distance Between Point P and Point Q We need to find the distance between $P(6, 3)$ and $Q(-2, -3)$. Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$: Let $(x_1, y_1) = (6, 3)$ and $(x_2, y_2) = (-2, -3)$. $PQ = \sqrt{(-2 - 6)^2 + (-3 - 3)^2}$ $PQ = \sqrt{(-8)^2 + (-6)^2}$ $PQ = \sqrt{64 + 36}$ $PQ = \sqrt{100}$ $PQ = 10$

Common Mistakes & Tips

• Incorrectly factoring exponential equations: Always factor out the smallest power of the base ($2^n$ in this case) to simplify the equation.
• Sign errors in the distance formula: Pay close attention to the signs of the coordinates when calculating the differences $(x_2 - x_1)$ and $(y_2 - y_1)$. Squaring negative numbers correctly is crucial.
• Assuming $m$ and $n$ are interchangeable: The problem statement implies $m$ and $n$ are specific to sets $A$ and $B$ respectively, and the difference $2^m - 2^n$ being positive means $m > n$.

Summary We translated the problem's condition about the number of subsets into an exponential equation $2^m - 2^n = 56$. By factoring out $2^n$ and using the prime factorization of 56, we determined that $n=3$ and $m-n=3$, leading to $m=6$. This gave us the coordinates of point $P$ as $(6, 3)$. Finally, we applied the distance formula to find the distance between $P(6, 3)$ and $Q(-2, -3)$, which is 10. This corresponds to option (A).

The final answer is $\boxed{10}$.