Two finite sets have m and `n ( m gt n)` elements .The number of subes of the first set is 112 more than that of the second set. The value of mn is

Options

A. 18
B. 28
C. 32
D. 36

Explanation

To solve the problem, we need to find the values of \( m \) and \( n \) given the condition about the number of subsets of two finite sets. Let's break down the solution step by step.

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

### Step 2: Set up the equation
According to the problem, the number of subsets of the first set is 112 more than that of the second set. This gives us the equation:
\[ 2^m = 2^n + 112 \]

### Step 3: Rearrange the equation
We can rearrange the equation to isolate \( 2^m \):
\[ 2^m - 2^n = 112 \]

### Step 4: Factor the left-hand side
We can factor the left-hand side:
\[ 2^n(2^{m-n} - 1) = 112 \]

### Step 5: Analyze the factors of 112
Now, we need to find pairs of \( (2^n, 2^{m-n} - 1) \) that multiply to give 112. The factors of 112 are:
- \( 1 \times 112 \)
- \( 2 \times 56 \)
- \( 4 \times 28 \)
- \( 7 \times 16 \)
- \( 8 \times 14 \)

Since \( 2^n \) must be a power of 2, we can only consider the factors \( 8 \) and \( 16 \) since \( 2^3 = 8 \) and \( 2^4 = 16 \).

### Step 6: Solve for \( n \) and \( m \)
1. **If \( 2^n = 8 \) (i.e., \( n = 3 \))**:
\[ 2^{m-n} - 1 = 14 \implies 2^{m-n} = 15 \quad \text{(not a power of 2)} \]

2. **If \( 2^n = 16 \) (i.e., \( n = 4 \))**:
\[ 2^{m-n} - 1 = 7 \implies 2^{m-n} = 8 \implies m-n = 3 \implies m = n + 3 = 4 + 3 = 7 \]

### Step 7: Calculate \( mn \)
Now that we have \( m = 7 \) and \( n = 4 \), we can find \( mn \):
\[ mn = 7 \times 4 = 28 \]

### Final Answer
Thus, the value of \( mn \) is \( 28 \).

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