Let's solve the problem step by step, using combinatorics. The key idea is to use the concept of combinations (for selecting people) and permutations (for arranging them).

a) The bride must be in the picture

We need to arrange 6 people in a row, where one of them is the bride. This means we need to select 5 more people from the remaining 9 (including the groom).

• Step 1: Select 5 people from the remaining 9 people (groom included). $\text{Number of ways to choose 5 people from 9} = \binom{9}{5}$

• Step 2: Arrange the 6 people (bride + 5 selected people). $\text{Number of ways to arrange 6 people} = 6!$

Thus, the total number of ways is: $\binom{9}{5} \times 6!$

b) Both the bride and groom must be in the picture

Now, both the bride and the groom must be in the picture, and we need to select 4 more people from the remaining 8 people.

• Step 1: Select 4 people from the remaining 8. $\text{Number of ways to choose 4 people from 8} = \binom{8}{4}$

• Step 2: Arrange the 6 people (bride, groom, and 4 selected people). $\text{Number of ways to arrange 6 people} = 6!$

Thus, the total number of ways is: $\binom{8}{4} \times 6!$

c) Exactly one of the bride and groom is in the picture

Here, we need to account for two cases:

• Case 1: The bride is in the picture, but the groom is not.
• Case 2: The groom is in the picture, but the bride is not.

In either case, we need to select 5 more people from the remaining 8 people.

• Step 1: Choose 5 people from the remaining 8 (excluding the other spouse). $\text{Number of ways to choose 5 people from 8} = \binom{8}{5}$

• Step 2: Arrange the 6 people (one of the bride or groom + 5 selected people). $\text{Number of ways to arrange 6 people} = 6!$

Since there are two cases (either the bride or the groom is in the picture), the total number of ways is: $2 \times \left( \binom{8}{5} \times 6! \right)$

Summary of the results:

• a) The total number of ways is $\binom{9}{5} \times 6!$.
• b) The total number of ways is $\binom{8}{4} \times 6!$.
• c) The total number of ways is $2 \times \left( \binom{8}{5} \times 6! \right)$.

Would you like me to calculate these expressions for you, or would you like to see more details on any part?

---

Related Questions:

1. How would the solution change if the bride or groom must be at a specific position?
2. What happens if only 4 people need to be arranged, including the bride and groom?
3. How many ways are there to select a group of people if there are restrictions on who can be together?
4. Can you arrange 6 people if the bride and groom can't stand next to each other?
5. What if only the bride is excluded from the selection?

Tip:

For complex combinatorics problems, break them down into simple steps—first handle the selection, then the arrangement.