In m, n , p are in A.P. and `m^n = p^m = n! + p, m, n, p in N`, then the value of m.n.p is, where `2 < m, n < p < 10` :

Options

A. 136
B. 72
C. 162
D. none of these

Answer

To solve the problem, we need to find the values of \( m, n, p \) such that they are in Arithmetic Progression (A.P.) and satisfy the equations \( m^n = p^m = n! + p \) with the constraints \( 2 < m, n < p < 10 \).

### Step-by-step Solution:

1. **Understanding A.P.**:
Since \( m, n, p \) are in A.P., we can express \( n \) as the average of \( m \) and \( p \):
\[ n = \frac{m + p}{2} \]

2. **Setting Up the Equations**:
From the problem statement, we have:
\[ m^n = p^m = n! + p \]

3. **Trial and Error Method**:
Given the constraints \( 2 < m, n < p < 10 \), we will try different values for \( m \) and \( p \) and calculate \( n \) accordingly.

4. **Trying Values**:
Let's start with \( m = 3 \) and \( p = 9 \):
- Calculate \( n \):
\[ n = \frac{3 + 9}{2} = 6 \]
- Check if \( m, n, p \) are in the required range:
\[ 2 < 3 < 6 < 9 < 10 \quad \text{(True)} \]

5. **Calculating \( m^n \) and \( p^m \)**:
- Calculate \( m^n \):
\[ m^n = 3^6 = 729 \]
- Calculate \( p^m \):
\[ p^m = 9^3 = 729 \]

6. **Calculating \( n! + p \)**:
- Calculate \( n! \):
\[ n! = 6! = 720 \]
- Calculate \( n! + p \):
\[ n! + p = 720 + 9 = 729 \]

7. **Verifying the Equations**:
We have:
\[ m^n = 729, \quad p^m = 729, \quad n! + p = 729 \]
All three expressions are equal, confirming our values are correct.

8. **Calculating \( m \cdot n \cdot p \)**:
Now we need to find \( m \cdot n \cdot p \):
\[ m \cdot n \cdot p = 3 \cdot 6 \cdot 9 \]
- Calculate:
\[ 3 \cdot 6 = 18 \]
\[ 18 \cdot 9 = 162 \]

### Final Answer:
Thus, the value of \( m \cdot n \cdot p \) is \( \boxed{162} \).