The number of ways in which ten candidates A1, A2, .....A10 can be ranked such that A1 is always above A10 is

This problem can be solved using the concept of permutations in combinatorics.

Step 1: Consider A1 and A10 as a single entity. Now, we have 9 entities in total (8 individual candidates and 1 entity consisting of A1 and A10).

Step 2: These 9 entities can be arranged in 9! (factorial) ways.

Step 3: However, within the entity of A1 and A10, A1 can be above A10 in only one way (since it's given that A1 is always above A10).

Step 4: Therefore, the total number of ways in which the ten candidates can be ranked such that A1 is always above A10 is 9! * 1 = 9!.

So, the answer is 9!.