for b) explain you why you don’t include a 2! since there’s clealry two sets made. 

You are correct to notice that in some permutation problems, there are multiple groups or sets that need to be accounted for. Let's clarify why a \(2!\) factor is not included in part (b) of this specific problem.

### Part (b) Detailed Explanation:

The problem asks us to determine the total possible number of arrangements where the 3 athletes from School A are in 3 lanes next to each other. 

To solve this, we consider the block of 3 athletes from School A as a single unit. Here’s the detailed step-by-step explanation:

1. **Treat the 3 Athletes from School A as One Unit:**
   - If we consider the 3 athletes as one single block, we have this block plus the 5 other athletes to arrange.
   - Thus, we have 6 units to arrange: the block of 3 athletes from School A and the other 5 athletes.

2. **Permutations of 6 Units:**
   - The number of ways to arrange 6 units (one block and 5 individuals) is \(6!\):
     \[
     6! = 720
     \]

3. **Arrangements Within the Block:**
   - Within the block, the 3 athletes from School A can be arranged among themselves in \(3!\) ways:
     \[
     3! = 6
     \]

Combining these two, the total number of arrangements where the 3 athletes from School A are together is:
\[
6! \times 3! = 720 \times 6 = 4320
\]

### Why Not Include \(2!\)?

The \(2!\) would be included if we had two identical groups or sets that could switch places, but that's not the case here. Each runner is unique, and there is no swapping of identical groups.

Here's an example when \(2!\) would be included:
- Suppose we had two identical blocks of athletes that could swap places. For instance, if there were two groups of three identical athletes (e.g., three from School A and three from School B) and we treated each group as a single unit, then the possible arrangements of these groups would involve a \(2!\) factor to account for swapping these identical groups.

In this problem, since we are only considering one block (the three athletes from School A as a single unit) and no identical groups that could switch places, we do not need to include \(2!\) in our calculations.

Would you like more details or have any questions about this explanation? Here are 5 questions you might ask next:

1. How do you handle problems with multiple identical groups or sets?
2. Can you provide an example where \(2!\) is included in permutation calculations?
3. What is the difference between permutations and combinations in problems like these?
4. How would the problem change if there were 4 athletes from School A?
5. Are there any shortcuts or formulas for solving more complex permutation problems?

**Tip:** Always break down permutation and combination problems into smaller parts, and visualize the arrangement to ensure you account for all unique arrangements and avoid overcounting identical sets.

Understand why the factorial 2! is not included in a specific permutation problem involving grouping of athletes. Detailed step-by-step explanation of permutations and factorial calculations. Suitable for high school students and anyone interested in combinatorics and permutation theory.

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