Consider the set `A= {3, 4, 5}` and the number of null relations, identity relation, universal relations,reflexive relation on A are respectively `n_1,n_2,n_3,n_4`,Then the value of `n_1+n_2+n_3+n_4` is equal to

Options

A. 8
B. 7
C. 73
D. 67

Answer

To solve the problem, we need to find the values of \( n_1 \) (number of null relations), \( n_2 \) (number of identity relations), \( n_3 \) (number of universal relations), and \( n_4 \) (number of reflexive relations) for the set \( A = \{3, 4, 5\} \).

### Step-by-Step Solution:

1. **Finding \( n_1 \) (Number of Null Relations)**:
- A null relation is a relation where no elements are related to each other.
- For any set with elements, there is only one null relation, which is the empty relation.
- Therefore, \( n_1 = 1 \).

2. **Finding \( n_2 \) (Number of Identity Relations)**:
- An identity relation on a set is a relation where each element is related to itself.
- For the set \( A = \{3, 4, 5\} \), the identity relation is \( \{(3,3), (4,4), (5,5)\} \).
- There is only one identity relation for any set, so \( n_2 = 1 \).

3. **Finding \( n_3 \) (Number of Universal Relations)**:
- A universal relation on a set is a relation where every element is related to every other element, including itself.
- For the set \( A \) with 3 elements, the universal relation includes all possible pairs of elements.
- The number of pairs is \( n^2 \) where \( n \) is the number of elements in the set. Thus, \( n_3 = 2^{n^2} = 2^{3^2} = 2^9 = 512 \).

4. **Finding \( n_4 \) (Number of Reflexive Relations)**:
- A reflexive relation on a set must include all pairs of the form \( (a, a) \) for each element \( a \) in the set.
- For the set \( A \) with 3 elements, we must include \( (3,3), (4,4), (5,5) \).
- The remaining pairs can either be included or not included, which gives us \( 2^{(n^2 - n)} \) options for the remaining pairs.
- Here, \( n^2 - n = 9 - 3 = 6 \), so \( n_4 = 2^6 = 64 \).

5. **Calculating the Total**:
- Now we sum all the values:
\[ n_1 + n_2 + n_3 + n_4 = 1 + 1 + 512 + 64 = 578 \]

### Final Answer:
The value of \( n_1 + n_2 + n_3 + n_4 \) is \( 578 \).