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!.

Question

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!.

Knowee AI · Accepted Answer

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!.

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