Find the number of ways in which 10 condidates `A_(1),A_(2),......,A_(10)` can be ranked so that `A_(1)` is always above `A_(2)`,

Options

A. 10!
B. `(10!)/(2)`
C. 9!
D. None of these

Explanation

To find the number of ways in which 10 candidates \( A_1, A_2, \ldots, A_{10} \) can be ranked such that \( A_1 \) is always above \( A_2 \), we can follow these steps:

### Step 1: Total Candidates
We have a total of 10 candidates, which we can denote as \( A_1, A_2, A_3, \ldots, A_{10} \).

**Hint:** Identify the total number of candidates involved in the ranking.

### Step 2: Condition on Ranking
The condition states that \( A_1 \) must always be ranked above \( A_2 \). This means in any valid ranking, \( A_1 \) should appear before \( A_2 \).

**Hint:** Understand the implications of the condition on the arrangement of candidates.

### Step 3: Choosing Positions for \( A_1 \) and \( A_2 \)
We can select 2 positions out of the 10 available positions for \( A_1 \) and \( A_2 \). The number of ways to choose 2 positions from 10 is given by the combination formula \( \binom{10}{2} \).

**Hint:** Use the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \) to find the number of ways to choose positions.

### Step 4: Arranging Remaining Candidates
After selecting positions for \( A_1 \) and \( A_2 \), we have 8 remaining candidates \( A_3, A_4, \ldots, A_{10} \). These 8 candidates can be arranged in the remaining 8 positions in \( 8! \) ways.

**Hint:** Remember that the arrangement of the remaining candidates is independent of the positions chosen for \( A_1 \) and \( A_2 \).

### Step 5: Total Arrangements
The total number of arrangements where \( A_1 \) is above \( A_2 \) can be calculated as:
\[ \text{Total Ways} = \binom{10}{2} \times 8! \]

### Step 6: Calculate \( \binom{10}{2} \)
Using the combination formula:
\[ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10!}{2! \times 8!} = \frac{10 \times 9}{2 \times 1} = 45 \]

**Hint:** Simplify the combination calculation to find the value.

### Step 7: Calculate Total Ways
Now substituting back into the total arrangements:
\[ \text{Total Ways} = 45 \times 8! \]

### Step 8: Calculate \( 8! \)
Calculating \( 8! \):
\[ 8! = 40320 \]

### Step 9: Final Calculation
Now, multiply the two results:
\[ \text{Total Ways} = 45 \times 40320 = 1814400 \]

### Conclusion
Thus, the total number of ways in which the 10 candidates can be ranked such that \( A_1 \) is always above \( A_2 \) is \( 1814400 \).

**Final Answer:** \( 1814400 \)