Let's analyze the three parts of the problem:

Given Set of Letters: A, B, C, D, E, F

(a) A and B are next to each other

If A and B must be next to each other, we can treat "AB" as a single unit or block. So, instead of arranging 6 letters, we are arranging 5 units: "AB", C, D, E, F.

1. Number of arrangements of 5 units:
   The number of ways to arrange these 5 units is $5!$.

2. Arrangement within the block "AB":
   A and B can be arranged within their block as "AB" or "BA". So, for each arrangement of the other 5 units, there are 2 arrangements of A and B.

Thus, the total number of linear arrangements where A and B are next to each other is: $5! \times 2 = 120 \times 2 = 240$

(b) A is before B

If A must be before B, we can consider the total number of possible arrangements of A, B, C, D, E, F without any restrictions, and then determine the fraction of these in which A is before B.

1. Total number of arrangements of all 6 letters:
   This is given by $6!$.

2. Arrangements where A is before B:
   In exactly half of these arrangements, A will be before B. This is because for any arrangement, either A is before B or B is before A, and both cases are equally likely.

So, the total number of arrangements where A is before B is: $\frac{6!}{2} = \frac{720}{2} = 360$

(c) A is before B, and B is before C

Here, A must be before B, and B must be before C.

1. Total number of arrangements of all 6 letters:
   Again, the total number of arrangements is $6!$.

2. Arrangements where A is before B, and B is before C:
   The letters A, B, and C can appear in any of the 6 positions, but we need A to be before B and B to be before C. Out of the $3!$ (6) possible arrangements of A, B, and C, only 1 has A before B and B before C.

Thus, the total number of arrangements where A is before B, and B is before C is: $\frac{6!}{3!} = \frac{720}{6} = 120$

Summary of Answers

1. (a) A and B are next to each other: $240$ arrangements.
2. (b) A is before B: $360$ arrangements.
3. (c) A is before B, and B is before C: $120$ arrangements.

Would you like details on any of these calculations, or have any questions?

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Here are 5 related questions to deepen your understanding:

1. How many different arrangements are there of the letters A, B, C, D, E, F where A is not next to B?
2. If the letters A and B are to be separated by exactly one letter, how many arrangements are possible?
3. How many arrangements are possible if A, B, and C must be next to each other in any order?
4. What if A, B, and C must be next to each other in the exact order A-B-C?
5. How many arrangements are possible if A and F are at the two ends of the arrangement?

Tip: For problems involving ordering or arrangements, consider breaking them down by treating groups of items as a single unit or by considering fractions of all possible outcomes.