1. How many ways can a photographer in a wedding arrange 5 people in a row from a group of 10
people, where the bride and the groom are among these 10 people, if
(a) the bride must be in the picture?
(b) both the bride and groom must be in the picture?
(c) exactly one of the bride or the groom is in the picture?

Let's solve each part step by step:

### (a) The bride must be in the picture
We have 10 people in total, including the bride. Since the bride must be in the picture, we choose 4 additional people from the remaining 9 people to fill the other spots in the arrangement. Then, we arrange the 5 people (including the bride) in a row.

1. **Choose 4 people from the remaining 9**:  
   This can be done in \( \binom{9}{4} \) ways.
   \[
   \binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126
   \]

2. **Arrange the 5 people in a row**:  
   This can be done in \( 5! \) ways.
   \[
   5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
   \]

Thus, the total number of ways to arrange the 5 people (with the bride) is:
\[
\binom{9}{4} \times 5! = 126 \times 120 = 15,120
\]

### (b) Both the bride and groom must be in the picture
If both the bride and groom must be in the picture, then we need to choose 3 more people from the remaining 8 people and arrange the 5 people in a row.

1. **Choose 3 people from the remaining 8**:  
   This can be done in \( \binom{8}{3} \) ways.
   \[
   \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56
   \]

2. **Arrange the 5 people in a row**:  
   This can be done in \( 5! \) ways.
   \[
   5! = 120
   \]

Thus, the total number of ways to arrange the 5 people (with both the bride and groom) is:
\[
\binom{8}{3} \times 5! = 56 \times 120 = 6,720
\]

### (c) Exactly one of the bride or the groom is in the picture
For this case, we have two sub-cases:
1. **The bride is in the picture, but not the groom.**
2. **The groom is in the picture, but not the bride.**

The number of ways to arrange each sub-case will be the same, so we calculate one and double it.

1. **Choose 4 people from the remaining 8 people (since either the bride or the groom is already in the picture)**:  
   This can be done in \( \binom{8}{4} \) ways.
   \[
   \binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70
   \]

2. **Arrange the 5 people in a row**:  
   This can be done in \( 5! = 120 \) ways.

Thus, for one sub-case (bride in, groom out or vice versa), the total number of arrangements is:
\[
\binom{8}{4} \times 5! = 70 \times 120 = 8,400
\]

Since either the bride or the groom is in the picture (but not both), we multiply by 2:
\[
2 \times 8,400 = 16,800
\]

### Final answers:
- **(a)**: 15,120 ways
- **(b)**: 6,720 ways
- **(c)**: 16,800 ways

Would you like more details on any of these steps or clarifications?

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Here are some related questions for further practice:

1. How many ways can you arrange 5 people in a row from a group of 10 without any restrictions?
2. How many ways can you arrange 4 people in a row from a group of 8?
3. If the bride must always be on one of the ends of the row, how many arrangements are possible in part (a)?
4. What if the bride and groom must stand next to each other? How does this change the answer for part (b)?
5. How would the solution change if the people were arranged in a circle instead of a row?

Tip: Remember that the order in which people are arranged matters in permutation problems, so multiplying by factorials is crucial!

How many ways can a photographer in a wedding arrange 5 people in a row from a group of 10 people, where the bride and the groom are among these 10 people, if (a) the bride must be in the picture? (b) both the bride and groom must be in the picture? (c) exactly one of the bride or the groom is in the picture?

Explore how to calculate the number of ways to arrange 5 people from a group of 10, including specific conditions for the bride and groom. This problem involves key concepts in combinatorics such as combinations and permutations. Step-by-step solutions included, suitable for Grades 10-12.

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