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

Options

A. `m=7,n=6`
B. `m=6,n=3`
C. `m=5,n=1`
D. `m=8,n=7`

Answer

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

### Step-by-step Solution:

1. **Understanding the Number of Subsets**:
The number of subsets of a set with \( k \) elements is given by \( 2^k \). Therefore, for the first set with \( m \) elements, the number of subsets is \( 2^m \), and for the second set with \( n \) elements, the number of subsets is \( 2^n \).

2. **Setting Up the Equation**:
According to the problem, we have:
\[ 2^m = 2^n + 56 \]

3. **Rearranging the Equation**:
We can rearrange the equation to express it in a more manageable form:
\[ 2^m - 2^n = 56 \]

4. **Factoring the Left Side**:
We can factor the left side:
\[ 2^n (2^{m-n} - 1) = 56 \]

5. **Finding Possible Values**:
Since \( 56 = 2^3 \times 7 \), we can set \( 2^n \) to be one of the factors of 56. The possible values for \( 2^n \) are \( 1, 2, 4, 8, 16, 32 \).

- If \( 2^n = 8 \) (which means \( n = 3 \)):
\[ 2^n (2^{m-n} - 1) = 8(2^{m-3} - 1) = 56 \]
\[ 2^{m-3} - 1 = 7 \implies 2^{m-3} = 8 \implies m - 3 = 3 \implies m = 6 \]

6. **Conclusion**:
The values of \( m \) and \( n \) are \( m = 6 \) and \( n = 3 \).

### Final Answer:
The values of \( m \) and \( n \) are \( m = 6 \) and \( n = 3 \).