Let A = {1,2,3,…..10} and `f:ArarrA` be defined as `f(k)={{:(k+1," if k is odd"),(k," if k is even"):}`
Then the number of possible functions g:`ArarrA` such that gof=f is

Options

A. `10^5`
B. `""^(10)C_5`
C. `5^5`
D. `5!`

Answer

To solve the problem, we need to find the number of functions \( g: A \to A \) such that \( g \circ f = f \), where \( f \) is defined as follows:

\[ f(k) = \begin{cases} k + 1 & \text{if } k \text{ is odd} \\ k & \text{if } k \text{ is even} \end{cases} \]

### Step-by-step Solution:

1. **Understand the Function \( f \)**:
- For odd \( k \): \( f(k) = k + 1 \)
- For even \( k \): \( f(k) = k \)
- The elements of set \( A = \{1, 2, 3, \ldots, 10\} \) include 5 odd numbers (1, 3, 5, 7, 9) and 5 even numbers (2, 4, 6, 8, 10).

2. **Determine the Output of \( f \)**:
- For odd \( k \):
- \( f(1) = 2 \)
- \( f(3) = 4 \)
- \( f(5) = 6 \)
- \( f(7) = 8 \)
- \( f(9) = 10 \)
- For even \( k \):
- \( f(2) = 2 \)
- \( f(4) = 4 \)
- \( f(6) = 6 \)
- \( f(8) = 8 \)
- \( f(10) = 10 \)

3. **Set Up the Condition \( g(f(k)) = f(k) \)**:
- If \( k \) is even (2, 4, 6, 8, 10), then \( f(k) = k \). Thus, \( g(k) \) must equal \( k \).
- If \( k \) is odd (1, 3, 5, 7, 9), then \( f(k) = k + 1 \). Thus, \( g(k + 1) \) must equal \( k + 1 \).

4. **Determine Values of \( g \)**:
- For \( k = 2, 4, 6, 8, 10\): \( g(2) = 2, g(4) = 4, g(6) = 6, g(8) = 8, g(10) = 10 \) (5 fixed points).
- For \( k = 1, 3, 5, 7, 9\): \( g(k + 1) \) can take any value from the set \( A \). Therefore:
- \( g(2) \) can be any of the 10 values.
- \( g(4) \) can be any of the 10 values.
- \( g(6) \) can be any of the 10 values.
- \( g(8) \) can be any of the 10 values.
- \( g(10) \) can be any of the 10 values.

5. **Calculate Total Number of Functions \( g \)**:
- Since \( g(k + 1) \) can take any value for each of the 5 odd \( k \):
- The total number of functions \( g \) is \( 10^5 \) (since there are 5 odd numbers).

### Final Answer:
The number of possible functions \( g: A \to A \) such that \( g \circ f = f \) is \( 10^5 \).