Problem: Two finite sets have m and n elements. The total no. of subsets of the first set is 56 more than the total no. of subsets of second set. Find the value of m and n.

Solution:
Let's start by understanding the concept of subsets. A subset is a set that contains elements of another set, including the empty set and the set itself. For example, the set {1, 2} has 4 subsets: { }, {1}, {2}, and {1, 2}.

We can use the formula to find the number of subsets of a set with n elements, which is 2^n. Using this formula, we can set up an equation to solve for m and n.

Let the two sets be A and B, where |A| = m and |B| = n.

The number of subsets of A is 2^m, and the number of subsets of B is 2^n.

We are given that the number of subsets of A is 56 more than the number of subsets of B. So we can set up the equation:

2^m = 2^n + 56

We can simplify this equation by taking the logarithm of both sides:

m log 2 = n log 2 + log 2^56

m log 2 = n log 2 + 56

We know that log 2 = 0.301, so we can substitute:

0.301m = 0.301n + 16.856

m = n + 56.012

Since m and n are both integers, we can use trial and error to find values that satisfy this equation. We can start with n = 1 and keep increasing n until we find a value that works.

When n = 5, m = 61, and 2^m - 2^n = 2^61 - 2^5 = 2,251,799,696, which is 56 more than 2^5 = 32.

Therefore, the solution is m = 61 and n = 5.

Let A and B be two sets having m& n elements respectively. According to question total no. of subset of A = 56 + total no. of subset of B. 2^m = 2^n+56. 2^m - 2^n = 56 2^n(2^m-n -1)= (8×7). 2^n(2^m-n -1)= 2^3 ×( 2^3 -1) on comparing both side. 2^n = 2^3 & 2^m-n =2^3. n = 3. & m-n = 3. m-3 = 3. m=6.