We’re missing a crucial part of your question: the conditions after the word “if”. From the sources this problem almost always comes with three sub‑questions like:

In how many ways can a photographer at a wedding arrange 6 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 and the groom is in the picture?

Below I’ll answer that standard version. If your exact “if …” is different, reply with the full sentence and I’ll adapt the answer.

In how many ways can a photographer at a wedding arrange 6 people in a row

from 10?

We have 10 people total, including a bride (B) and groom (G).
We always arrange 6 people in a row (ordered; so this is permutations).

a) Bride must be in the picture

Step 1: Choose who is in the photo.

  • The bride is definitely in.
  • We need 5 more people from the remaining 9 (everyone except B).
  • Number of ways to choose those 5 people:
    (95)\binom{9}{5}(59​).

Step 2: Arrange the chosen 6 in a row.

  • Any ordering of these 6 is allowed.
  • Number of arrangements: 6!6!6!.

Total ways:

(95)⋅6!=126⋅720=90,720.\binom{9}{5}\cdot 6!=126\cdot 720=90{,}720.(59​)⋅6!=126⋅720=90,720.

So, there are 90,720 ways when the bride must be in the picture.

b) Both bride and groom must be in the picture

Step 1: Choose who is in the photo.

  • B and G are definitely in.
  • We need 4 more people from the remaining 8.
  • Number of ways to choose those 4 people:
    (84)\binom{8}{4}(48​).

Step 2: Arrange the chosen 6 in a row.

  • Again, any ordering allowed.
  • Number of arrangements: 6!6!6!.

Total ways:

(84)⋅6!=70⋅720=50,400.\binom{8}{4}\cdot 6!=70\cdot 720=50{,}400.(48​)⋅6!=70⋅720=50,400.

So, there are 50,400 ways when both bride and groom must be in the picture.

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

Here we want photos that contain either B or G, but not both. Think of it as two symmetric cases:

  1. Bride in, groom out.
  2. Groom in, bride out.

Each case is counted the same way.

Case 1: Bride in, groom out

Step 1: Choose who is in the photo.

  • B is in, G is out.
  • So we choose 5 more people from the remaining 8 (everyone except B and G).
  • Number of ways: (85)\binom{8}{5}(58​).

Step 2: Arrange the chosen 6 in a row.

  • Number of arrangements: 6!6!6!.

Ways for this case:

(85)⋅6!=56⋅720=40,320.\binom{8}{5}\cdot 6!=56\cdot 720=40{,}320.(58​)⋅6!=56⋅720=40,320.

Case 2: Groom in, bride out

By symmetry, this is the same count:

40,320.40{,}320.40,320.

Total ways for “exactly one of B and G”:

2⋅40,320=80,640.2\cdot 40{,}320=80{,}640.2⋅40,320=80,640.

So, there are 80,640 ways when exactly one of the bride and groom is in the picture.

Compact formula summary (HTML table)

Here is a quick reference table:

[3][7] [8][3] [7][9]
Condition Counting setup Number of ways
Bride must be in the picture $$\binom{9}{5} \cdot 6!$$ 90,720
Both bride and groom must be in the picture $$\binom{8}{4} \cdot 6!$$ 50,400
Exactly one of bride or groom in the picture $$2 \cdot \binom{8}{5} \cdot 6!$$ 80,640
If your version has a different “if …” clause (for example, “bride next to groom”, “bride not next to groom”, etc.), paste the full text and I’ll solve that specific variant.