, the number of ways in which 10 candidates `A_(1)andA_(2)` can be ranked if if A1 is always above A2

Options

A. 9!
B. 2(9!)
C. `(1)/(2)(9!)`
D. none of these

Answer

To solve the problem of ranking 10 candidates \( A_1, A_2, A_3, \ldots, A_{10} \) with the condition that \( A_1 \) is always above \( A_2 \), we can follow these steps:

### Step 1: Calculate Total Arrangements
First, we need to find the total number of arrangements of the 10 candidates without any restrictions. The number of ways to arrange \( n \) candidates is given by \( n! \) (n factorial).

For our case, where \( n = 10 \):
\[ \text{Total arrangements} = 10! \]

### Step 2: Consider the Condition
Since we want \( A_1 \) to always be ranked above \( A_2 \), we can think about how many of the total arrangements meet this condition.

In any arrangement of the 10 candidates, \( A_1 \) and \( A_2 \) can either be in the order \( (A_1, A_2) \) or \( (A_2, A_1) \). These two arrangements are equally likely.

### Step 3: Divide Total Arrangements
Since half of the arrangements will have \( A_1 \) above \( A_2 \) and the other half will have \( A_2 \) above \( A_1 \), we can find the number of arrangements where \( A_1 \) is above \( A_2 \) by dividing the total arrangements by 2.

Thus, the number of arrangements where \( A_1 \) is above \( A_2 \) is:
\[ \text{Desired arrangements} = \frac{10!}{2} \]

### Step 4: Calculate the Final Answer
Now we can compute the final answer:
\[ 10! = 3628800 \]
\[ \text{Desired arrangements} = \frac{3628800}{2} = 1814400 \]

### Conclusion
Therefore, the number of ways in which the 10 candidates can be ranked such that \( A_1 \) is always above \( A_2 \) is:
\[ \boxed{1814400} \]