How many different linear arrangements are there of the letters A, B, C, D, E, F for which

(a) A and B are next to each other?

(b) A is before B?

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

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.

How many different linear arrangements are there of the letters A, B, C, D, E, F for which (a) A and B are next to each other? (b) A is before B? (c) A is before B and B is before C?

Explore how to calculate different linear arrangements of the letters A, B, C, D, E, F. This problem covers key combinatorics concepts such as permutations and factorials, and includes detailed solutions for cases where A and B are next to each other, A is before B, and A is before B and B before C. Suitable for students in Grades 10-12.

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