Solution:

Introduction:
In this problem, we have to find the number of ways in which speaker A will always speak before speaker B. There are a total of 5 speakers A, B, C, D, and E.

Approach:
To solve this problem, we can use the concept of permutations. We know that the number of ways to arrange n objects is n!. In this case, we have 5 speakers, so the total number of ways to arrange them is 5!. However, we want to find the number of ways in which A will always speak before B.

Case 1: A and B are adjacent
If we consider A and B as a single unit, then there are 4 units to arrange. This can be done in 4! ways. However, A and B can be arranged in two ways (AB or BA). Therefore, the number of ways in which A will always speak before B when they are adjacent is 2 x 4! = 48.

Case 2: A and B are not adjacent
If A and B are not adjacent, then we can fix the position of A and B. There are 4 positions where A can be placed: first, second, third, or fourth. Once A is fixed, there are 3 positions where B can be placed. The remaining 3 speakers can be arranged in 3! ways. Therefore, the number of ways in which A will always speak before B when they are not adjacent is 4 x 3 x 3! = 72.

Total number of ways:
The total number of ways in which A will always speak before B is the sum of the number of ways in the two cases above: 48 + 72 = 120.

Conclusion:
Therefore, there are 120 ways in which A will always speak before B.