Consider two sets A = {a, b}, B = {e, f}. If maximum numbers of relations from A to B, A to A, B to B are l, m, n respectively then the value of 2l - m - n is

Options

A. 8
B. 0
C. 16
D. 32

Answer

To solve the problem, we need to find the maximum number of relations from set A to set B, from set A to set A, and from set B to set B. We will denote these maximum numbers of relations as \( l \), \( m \), and \( n \) respectively.

### Step 1: Determine the number of elements in sets A and B
- Set \( A = \{a, b\} \) has 2 elements.
- Set \( B = \{e, f\} \) also has 2 elements.

### Step 2: Calculate \( l \) (relations from A to B)
The number of relations from set \( A \) to set \( B \) can be calculated using the formula:
\[ l = 2^{(n_A \times n_B)} \]
where \( n_A \) is the number of elements in set \( A \) and \( n_B \) is the number of elements in set \( B \).

Here, \( n_A = 2 \) and \( n_B = 2 \):
\[ l = 2^{(2 \times 2)} = 2^4 = 16 \]

### Step 3: Calculate \( m \) (relations from A to A)
The number of relations from set \( A \) to itself is given by:
\[ m = 2^{(n_A \times n_A)} \]
Using \( n_A = 2 \):
\[ m = 2^{(2 \times 2)} = 2^4 = 16 \]

### Step 4: Calculate \( n \) (relations from B to B)
Similarly, the number of relations from set \( B \) to itself is:
\[ n = 2^{(n_B \times n_B)} \]
Using \( n_B = 2 \):
\[ n = 2^{(2 \times 2)} = 2^4 = 16 \]

### Step 5: Calculate \( 2l - m - n \)
Now we can substitute the values of \( l \), \( m \), and \( n \) into the expression \( 2l - m - n \):
\[ 2l = 2 \times 16 = 32 \]
\[ 2l - m - n = 32 - 16 - 16 = 32 - 32 = 0 \]

### Final Answer
Thus, the value of \( 2l - m - n \) is \( 0 \).

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