Two finite sets have m and n elements respectively . The total number of subsets of first set is 56 more than the total number of subsets of the second. Find the values of m and n.

Answer

To solve the problem, we need to find the values of \( m \) and \( n \) given that the total number of subsets of the first set is 56 more than the total number of subsets of the second set.

Let's break down the solution step by step:

1. **Identify the number of subsets for each set:**
- Let set \( A \) have \( m \) elements.
- Let set \( B \) have \( n \) elements.
- The number of subsets of set \( A \) is \( 2^m \).
- The number of subsets of set \( B \) is \( 2^n \).

2. **Set up the equation based on the given condition:**
- According to the problem, the number of subsets of the first set exceeds the number of subsets of the second set by 56.
- This can be written as:
\[ 2^m - 2^n = 56 \]

3. **Rewrite the equation to factor out common terms:**
- We can rewrite \( 2^m \) as \( 2^n \cdot 2^{m-n} \):
\[ 2^n \cdot 2^{m-n} - 2^n = 56 \]
- Factor out \( 2^n \):
\[ 2^n (2^{m-n} - 1) = 56 \]

4. **Express 56 as a product of powers of 2 and another factor:**
- Notice that 56 can be written as \( 8 \times 7 \), and 8 is \( 2^3 \):
\[ 2^n (2^{m-n} - 1) = 2^3 \times 7 \]

5. **Compare the powers of 2 on both sides:**
- For the equation to hold, \( 2^n \) must be \( 2^3 \), so:
\[ 2^n = 2^3 \implies n = 3 \]

6. **Substitute \( n = 3 \) back into the equation:**
- Substitute \( n = 3 \) into the factored equation:
\[ 2^3 (2^{m-3} - 1) = 56 \]
- Simplify:
\[ 8 (2^{m-3} - 1) = 56 \]
- Divide both sides by 8:
\[ 2^{m-3} - 1 = 7 \]
- Add 1 to both sides:
\[ 2^{m-3} = 8 \]
- Since \( 8 = 2^3 \):
\[ 2^{m-3} = 2^3 \]

7. **Compare the exponents to solve for \( m \):**
- Since the bases are the same, the exponents must be equal:
\[ m - 3 = 3 \implies m = 6 \]

Therefore, the values of \( m \) and \( n \) are:
\[ m = 6 \quad \text{and} \quad n = 3 \]