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 seccond 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 step by step, we need to find the values of \( m \) and \( n \) such that the total number of subsets of the first set is 56 more than the total number of subsets of the second set.

### Step 1: Understand the formula for the number of subsets
The number of subsets of a set with \( m \) elements is given by \( 2^m \). Similarly, the number of subsets of a set with \( n \) elements is \( 2^n \).

### Step 2: Set up the equation based on the problem statement
According to the problem, the number of subsets of the first set is 56 more than that of the second set. Therefore, we can write the equation:
\[ 2^m = 2^n + 56 \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:
\[ 2^m - 2^n = 56 \]

### Step 4: Factor out \( 2^n \)
We can factor out \( 2^n \) from the left side:
\[ 2^n (2^{m-n} - 1) = 56 \]

### Step 5: Find the factors of 56
Now, we need to find pairs of \( (2^n, 2^{m-n} - 1) \) that multiply to give 56. The factors of 56 are:
- \( 1 \times 56 \)
- \( 2 \times 28 \)
- \( 4 \times 14 \)
- \( 7 \times 8 \)

### Step 6: Analyze the factors
Since \( 2^n \) must be a power of 2, we can only consider the pairs where the first factor is a power of 2:
- \( 2^n = 4 \) and \( 2^{m-n} - 1 = 14 \)
- \( 2^n = 8 \) and \( 2^{m-n} - 1 = 7 \)

### Step 7: Solve for \( n \) and \( m \)
1. For \( 2^n = 4 \):
- \( n = 2 \)
- \( 2^{m-n} - 1 = 14 \) implies \( 2^{m-2} = 15 \) (not a power of 2)

2. For \( 2^n = 8 \):
- \( n = 3 \)
- \( 2^{m-n} - 1 = 7 \) implies \( 2^{m-3} = 8 \) which gives \( m - 3 = 3 \) or \( m = 6 \)

### Final Values
Thus, we find:
- \( m = 6 \)
- \( n = 3 \)

### Conclusion
The values of \( m \) and \( n \) are \( 6 \) and \( 3 \) respectively.