Inviting four people to dinner if one of them can be invited only if that person's two friends will be there as well

Question

A woman has 9 friends and must invite 4 of them to a dinner. The question is divided in two parts;

a) How many choices does she have if one of her friends can't be together simultaneously with other two friends?

For this part I did: $\displaystyle {9\choose 4} - {3\choose 3} {6\choose 1}$

I found a lot of questions like this on the forum, so I hope I got this part right (but I'm not good with math, so I can't be sure). My problem is the second part.

b) How many choices does she have if one of her friends won't accept the invite unless that person's two friends are at the dinner as well.

I'm not sure how can I model this. I tried to list all the possibilities of this part.

1. This person and his 2 friends are selected, so we have ${6 \choose 1}$ people left to choose.
2. If one of these two friends will not be there, this person will not go as well. So we have $2\cdot {7\choose 4}$ choices (times 2 because it's a case for each friend).
3. If both friends will not be at the party then this person will not be there as well so we have $\binom{6}{4}$ people left to select.

Then I just sum all these 3 cases, but I feel like I am over counting the possibilities. Do I really need the third case?

Unfortunately, there is no answer for this question to check. Hope someone can help me.

Answer 1

For part $(a)$, I interpret the question as suppose person $A$ won't attendd if $B$ attend or $C$ attend, then

If $A$ attends, then we won't invite $B$ or $C$ and out of the remaining $6$ people, we get to choose $3$ of them.

If $A$ doesn't attend, then we get to choose any $4$ people out of the remaining $8$.

$$\binom63+\binom84.$$

For part $(b)$, again, consider the two cases,

If $A$ attends, then we must invite $B$ and $C$ and out of the remaining $6$ people, we get to choose $1$ of them.

If $A$ doesn't attend, then we get to choose any $4$ out of the remaining $8$.

$$\binom61+\binom84$$