Consider the function f defined on the set of all non-negative interger such that `f(0) = 1, f(1) =0` and `f(n) + f(n-1) = nf(n-1)+(n-1) f(n-2)` for `n ge 2`, then f(5) is equal to

Options

A. 40
B. 44
C. 45
D. 60

Explanation

To solve the problem, we need to compute the values of the function \( f(n) \) using the given recurrence relation and initial conditions.

### Step-by-Step Solution:

1. **Initial Conditions**: We are given:
- \( f(0) = 1 \)
- \( f(1) = 0 \)

2. **Recurrence Relation**: The recurrence relation is given as:
\[ f(n) + f(n-1) = n f(n-1) + (n-1) f(n-2) \quad \text{for } n \geq 2 \]
We can rearrange this to express \( f(n) \):
\[ f(n) = n f(n-1) + (n-1) f(n-2) - f(n-1) \]
Simplifying this gives:
\[ f(n) = (n-1) f(n-1) + (n-1) f(n-2) \]
Factoring out \( (n-1) \):
\[ f(n) = (n-1)(f(n-1) + f(n-2)) \]

3. **Calculate \( f(2) \)**:
- Substitute \( n = 2 \):
\[ f(2) = (2-1)(f(1) + f(0)) = 1 \cdot (0 + 1) = 1 \]

4. **Calculate \( f(3) \)**:
- Substitute \( n = 3 \):
\[ f(3) = (3-1)(f(2) + f(1)) = 2 \cdot (1 + 0) = 2 \]

5. **Calculate \( f(4) \)**:
- Substitute \( n = 4 \):
\[ f(4) = (4-1)(f(3) + f(2)) = 3 \cdot (2 + 1) = 3 \cdot 3 = 9 \]

6. **Calculate \( f(5) \)**:
- Substitute \( n = 5 \):
\[ f(5) = (5-1)(f(4) + f(3)) = 4 \cdot (9 + 2) = 4 \cdot 11 = 44 \]

Thus, the value of \( f(5) \) is \( 44 \).

### Final Answer:
\[ f(5) = 44 \]