Counting Problem With Distinct Family Members For a Photo

Question

In this family there are 10 members all together, and 2 of the 10 are the parents. A photographer wants to get a photo of 6 members out of 10 in a row for the picture. How many ways can the photographer arrange the members if:

1. The oldest parent must be in the picture.

   My Solution: My thinking on this is that we choose the oldest parent, and we just multiply the other members in subtracting 1. $$1\cdot9\cdot8\cdot7\cdot6\cdot5$$

2. Both parents are in the picture.

   My Solution: Putting the two parents in the photo subtracts from the number of members that can be in the group. $$2\cdot8\cdot7\cdot6\cdot5\cdot4$$

3. Exactly one of the parents must be in the picture.

   My Solution: One parent is only allowed on the photo and the other parent cannot be the photo so: $$1\cdot8\cdot7\cdot6\cdot5\cdot4$$

I am wondering if this the right way to go about this problem. I've seen other problems use Inclusion-Exclusion, but I don't know if that applies here, or to be honest any other counting technique. Looking for some feedback on this problem seeing if my logic is correct or flawed.

Answer 1

A common beginner's mistake in combinatorics is trying to get everything into a single sequential product expression. This is fine in say, arranging $5$ objects out of $10$ but when applied more restraints, one has to take each process into account.

Here we have two processes namely, selection (subject to some restraints) followed by permutation. The answer should reflect both.

1. Choose oldest parent, choose 5 more, arrange all 6 $$ 1 \cdot \binom{9}{5} \cdot 6!$$ which is $6$ times your result as the oldest parent can stand in any of $6$ places.
2. Choose both parents, choose 4 more, arrange all 6 $$ 1 \cdot \binom{8}{4} \cdot 6!$$
3. Choose only one parent, choose 5 more, arrange all 6 $$ \binom{2}{1} \cdot \binom{8}{5} \cdot 6!$$

So making a map of processes is the way to approach.