Consider set `A = {x_(1), x_(2), x_(3), x_(4), x_(5)}` and set `B = {y_(1), y_(2), y_(3)}`. Function f is defined from A to B.
Number of function from A to B such that `f(x_(1)) = y_(1)` and `f(x_(2)) != y_(2)` is

Options

A. 26
B. 50
C. 14
D. 21

Explanation

To solve the problem, we need to find the number of functions \( f: A \to B \) that satisfy the conditions \( f(x_1) = y_1 \) and \( f(x_2) \neq y_2 \).

### Step-by-step Solution:

1. **Fix \( f(x_1) \)**:
Since we know that \( f(x_1) = y_1 \), we can start by fixing this value. This means we only need to assign values for \( x_2, x_3, x_4, \) and \( x_5 \).

2. **Count the remaining elements in set B**:
After assigning \( f(x_1) = y_1 \), the remaining elements in set \( B \) are \( \{y_2, y_3\} \).

3. **Assign \( f(x_2) \)**:
The condition states that \( f(x_2) \neq y_2 \). Therefore, \( f(x_2) \) can either be \( y_1 \) or \( y_3 \). Since \( f(x_1) = y_1 \), the only option left for \( f(x_2) \) is \( y_3 \).

4. **Assign values for \( x_3, x_4, \) and \( x_5 \)**:
Now, we have three elements \( x_3, x_4, \) and \( x_5 \) that can be assigned values from the set \( B \) which contains \( \{y_1, y_2, y_3\} \).

5. **Calculate the number of assignments**:
Each of \( x_3, x_4, \) and \( x_5 \) can take any of the three values in \( B \). Therefore, the number of ways to assign values to \( x_3, x_4, \) and \( x_5 \) is:
\[ 3 \times 3 \times 3 = 3^3 = 27 \]

6. **Final count**:
Since we have already determined that \( f(x_2) \) must be \( y_3 \), and the assignments for \( x_3, x_4, \) and \( x_5 \) can be done in 27 ways, the total number of functions from \( A \) to \( B \) that satisfy the given conditions is:
\[ 27 \]

### Conclusion:
The number of functions from \( A \) to \( B \) such that \( f(x_1) = y_1 \) and \( f(x_2) \neq y_2 \) is **27**.