For numbers $$ 2-6 $$, refer to the scenario below. Show your solutions on a sheer of paper. Refer to the criteria given in assessing your answer. A school is organizing a talent show. There are 8 participants, and the order in which they perform matters. The organizers want to determine the number of ways to arrange the performers, select a specific group for a special act, and explore other related scenarios involving permutations. How many different ways can the 8 participants be arranged in a performance lineup? If only 3 out of the 8 participants are chosen for a special act, and the order of performance matters, how many arrangements are possible? If two participants insist on performing one after the other, how many valid arrangements can be made for the 8 participants? If three participants are indistinguishable (e.g., they form a group act). how many unique arrangements can be made for all 8 participants? Suppose 2 out of the 8 participants are unavailable for the show. How many ways can the remaining 6 participants be arranged in the lineup? For numbers 7 -LI. refer to the scenario below. Show your solutions on a sheet of paper. Refer to the criteria given in

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Let's solve the problems step by step, focusing on permutations and arrangements of the participants.

### Problem 1: Arranging 8 Participants
To find the number of ways to arrange 8 participants, we use the formula for permutations of $$ n $$ distinct objects, which is given by $$ n! $$.

$$
n = 8 \implies 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
$$

Calculating $$ 8! $$:

$$
8! = 40320
$$

So, there are **40,320** different ways to arrange the 8 participants.

### Problem 2: Arranging 3 out of 8 Participants
When selecting and arranging 3 participants from 8, we use the permutation formula $$ P(n, r) = \frac{n!}{(n-r)!} $$.

Here, $$ n = 8 $$ and $$ r = 3 $$:

$$
P(8, 3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6
$$

Calculating $$ P(8, 3) $$:

$$
P(8, 3) = 8 \times 7 \times 6 = 336
$$

So, there are **336** arrangements possible for the special act.

### Problem 3: Two Participants Performing Together
If two participants must perform one after the other, we can treat them as a single unit or block. This reduces the number of units to arrange from 8 to 7 (the block plus the other 6 participants).

The number of arrangements is given by:

$$
7! \times 2!
$$

Where $$ 2! $$ accounts for the arrangements within the block.

Calculating:

$$
7! = 5040 \quad \text{and} \quad 2! = 2
$$

Thus, the total arrangements are:

$$
7! \times 2! = 5040 \times 2 = 10080
$$

So, there are **10,080** valid arrangements.

### Problem 4: Three Indistinguishable Participants
If three participants are indistinguishable, we can use the formula for permutations of multiset:

$$
\frac{n!}{n_1! \times n_2! \times \ldots}
$$

Here, $$ n = 8 $$ and $$ n_1 = 3 $$ (indistinguishable participants), and the rest are distinct:

$$
\frac{8!}{3!} = \frac{40320}{6} = 6720
$$

So, there are **6,720** unique arrangements.

### Problem 5: Arranging 6 Participants
If 2 out of the 8 participants are unavailable, we are left with 6 participants to arrange. The number of arrangements is simply:

$$
6! = 720
$$

So, there are **720** ways to arrange the remaining 6 participants.

### Summary of Results
1. **40,320** ways to arrange 8 participants.
2. **336** arrangements for 3 out of 8 participants.
3. **10,080** valid arrangements with two participants together.
4. **6,720** unique arrangements with three indistinguishable participants.
5. **720** ways to arrange the remaining 6 participants.
- **40,320** ways to arrange all 8 participants. - **336** arrangements for selecting and ordering 3 out of 8. - **10,080** valid arrangements when two participants must perform together. - **6,720** unique arrangements when three participants are indistinguishable. - **720** ways to arrange the remaining 6 participants after two are unavailable.

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