Animated Solution for Mathematics - Sets and Relations: Let $A$ and $B$ be two finite sets with $m$ and $n$ elements respectively. The total number of subsets of the set $A$ is 56 more than the total number of subsets of $B$. Then the distance of the point $P(m,n)$ from the point $Q(-2,-3)$ is

Welcome, future engineers! Today, we are going to embark on a journey that bridges two seemingly distant worlds: the discrete, logical realm of Set Theory and the continuous, visual world of Coordinate Geometry. It is a beautiful intersection, and I want you to see that math is not just about memorizing formulas; it is about finding the hidden patterns that connect different disciplines.

Let us start with the foundation. We have two finite sets, $A$ and $B$, with $m$ and $n$ elements respectively. The problem tells us that the number of subsets of $A$ is $56$ more than the number of subsets of $B$.

Recall the fundamental theorem: for any set with $k$ elements, the total number of subsets is $2^k$. This is a direct consequence of the fact that for each element, we have two choices: either include it in the subset or exclude it. Thus, we have the equation:

This is our starting point. It looks simple, but it is a Diophantine equation—an equation where we seek integer solutions. How do we solve for two variables with only one equation? The secret lies in the structure of the numbers themselves.

When you see an equation like $2^m-2^n=56$, your first instinct should be to factor. We have a common factor of $2^n$. Let us pull it out:

Now, look closely at this expression. We have a product of two terms: $2^n$ and $(2^{m-n}-1)$. The first term, $2^n$, is a power of $2$, which is inherently even (for $n\ge 1$).

The second term, $2^{m-n}-1$, is an odd number because any power of $2$ is even, and subtracting $1$ makes it odd. We have effectively split our number $56$ into an even part and an odd part. This is the "Aha!" moment.

Now, let us look at the right side of the equation: $56$. Let us perform a prime factorization. We know that $56=8\times 7$. And $8$ is simply $2^3$. So, we have:

By comparing our factored equation $2^n(2^{m-n}-1)=2^3\times 7$ with the prime factorization, we can equate the even parts and the odd parts. The even part $2^n$ must correspond to $2^3$, and the odd part $2^{m-n}-1$ must correspond to $7$. This is the elegance of the method!

Equating the powers of $2$, we get $n=3$. Equating the odd parts, we get $2^{m-n}-1=7$, which simplifies to $2^{m-n}=8$, or $2^{m-n}=2^3$. Thus, $m-n=3$. Since we already know $n=3$, it follows that $m=6$. We have found our coordinates!

Now that we have $m=6$ and $n=3$, our point $P(m,n)$ is simply $P(6,3)$. We are asked to find the distance of this point from $Q(-2,-3)$. Imagine the coordinate plane. We are calculating the length of the hypotenuse of a right-angled triangle formed by the horizontal and vertical differences between these two points.

Using the distance formula, $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$, we substitute our values:

Be careful with those negative signs! They are the most common trap in coordinate geometry. The expression becomes:

This is a classic $6-8-10$ triangle. Squaring the terms, we get $\sqrt{64+36}=\sqrt{100}=10$. The distance is exactly $10$ units.

We started with the abstract concept of subsets and ended with a concrete geometric distance. This is the beauty of mathematics. We didn't just calculate; we navigated through logic, factorization, and geometry. Keep this mindset—always look for the structure, always look for the connection—and you will find that even the most daunting JEE problems become a series of logical, satisfying steps.