Consider all function `f: {1,2,3,4} to {1,2,3,4}` which are one-one, onto and satisfy the following property :
If `f (k)` is odd then `f (k+1)` is even,` K=1,2,3.` The number of such function is :

Options

A. 4
B. 8
C. 12
D. 16

Answer

To solve the problem, we need to find the number of one-to-one and onto functions \( f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4\} \) that satisfy the condition: if \( f(k) \) is odd, then \( f(k+1) \) is even for \( k = 1, 2, 3 \).

### Step-by-Step Solution:

1. **Identify the elements**:
The set \( \{1, 2, 3, 4\} \) contains two odd numbers (1 and 3) and two even numbers (2 and 4).

2. **Understanding the condition**:
The condition states that if \( f(k) \) is odd, then \( f(k+1) \) must be even. This implies that:
- If \( f(1) \) is odd, then \( f(2) \) must be even.
- If \( f(2) \) is odd, then \( f(3) \) must be even.
- If \( f(3) \) is odd, then \( f(4) \) must be even.

3. **Case Analysis**:
We can analyze two cases based on whether \( f(1) \) is odd or even.

**Case 1**: \( f(1) \) is odd.
- Possible choices for \( f(1) \): 1 or 3 (2 choices).
- If \( f(1) \) is odd, \( f(2) \) must be even. Possible choices for \( f(2) \): 2 or 4 (2 choices).
- If \( f(2) \) is even, \( f(3) \) must be odd. The remaining odd number will be chosen for \( f(3) \) (1 choice).
- Finally, \( f(4) \) must take the remaining even number (1 choice).

Therefore, the total number of functions for this case is:
\[ 2 \text{ (choices for } f(1)) \times 2 \text{ (choices for } f(2)) \times 1 \text{ (choice for } f(3)) \times 1 \text{ (choice for } f(4)) = 4. \]

**Case 2**: \( f(1) \) is even.
- Possible choices for \( f(1) \): 2 or 4 (2 choices).
- If \( f(1) \) is even, \( f(2) \) can be odd. Possible choices for \( f(2) \): 1 or 3 (2 choices).
- If \( f(2) \) is odd, \( f(3) \) must be even. The remaining even number will be chosen for \( f(3) \) (1 choice).
- Finally, \( f(4) \) must take the remaining odd number (1 choice).

Therefore, the total number of functions for this case is:
\[ 2 \text{ (choices for } f(1)) \times 2 \text{ (choices for } f(2)) \times 1 \text{ (choice for } f(3)) \times 1 \text{ (choice for } f(4)) = 4. \]

4. **Total Count**:
Adding both cases together gives us:
\[ 4 \text{ (from Case 1)} + 4 \text{ (from Case 2)} = 8. \]

### Final Answer:
The total number of such functions is \( 8 \).