How should i go about answering this combination question?

Question

In how many ways can a coach select a team of $4$ players for table tennis if two of the players, $A$ and $B$ play well together and should either both play or both be excluded , and two others $X$ and $Y$ have recently quarreled and should not both be included? Nine players are available.

I have done it this way: 1. Combinations when both $A$ and $B$ are included but excluding $X$ and $Y$ together then adding combinations of not picking $A$ and $B$ still excluding picking both $X$ and $Y$: $$(_6C_2 . 1)2+(_6C_4)2 = 60$$ is this the right approach?

Answer 1

We consider two cases: either $\,A\,$ and $\,B\,$ are both on the team, or neither are.

Case 1: Both $\,A\,$ and $\,B\,$ are on the team. Then we have to choose the two remaining team members from the other seven players; however, we have to make sure that $\,X\,$ and $\,Y\,$ are not both on our team, and the only way that happens is in the single case $\,\{A,B,X,Y\}.\,$ Thus the number of possible teams is

$$\binom{7}{2} - 1 = 21 - 1 = 20.$$

Case 2: Neither $\,A\,$ nor $\,B\,$ are on the team. Then we have to choose all four team members from the other seven players. Once again we have to make sure that $\,X\,$ and $\,Y\,$ are not both on our team; this time, this will happen if only two of the team members are chosen from among the five players that are not $\,X\,$ or $\,Y.\,$ So in this case the number of possible teams is

$$\binom{7}{4} - \binom{5}{2} = 35 - 10 = 25.$$

So the total number of possible teams is

$$\left[\binom{7}{2} - 1\right] + \left[\binom{7}{4} - \binom{5}{2}\right] = 20 + 25 = \boxed{45}$$

Answer 2

We consider several cases.

1. The simplest case is when none of $A,B,X,Y$ is chosen. Then there are $5\choose 4$ possible teams.
2. When $A$ and $B$ are both chosen, there are $5\choose 2$ ways to select the other two players when none of $X$ and $Y$ is chosen, so there are $10$ possibilities where $X$ and $Y$ are both not chosen and $A$ and $B$ are both chosen.
3. There are two ways to select one of $X$ and $Y$ and $5\choose 3$ ways to select the rest of the team w/o $A$ and $B$. So there are $2\cdot 10=20$ ways in this case.
4. Finally, there are two ways to select one of $X$ and $Y$ and $5\choose 1$ ways to select the rest of the team with $A$ and $B$, for a total of $2\cdot 5=10$ ways for this case.

Thus there are $5+10+20+10=45$ ways in total.

Answer 3

Case 1: AB,X,${5\choose 1}=5$.

Case 2: AB,no X, ${6\choose 2}=15$.

Case 3: No AB, X, ${5\choose 3}=10$.

Case 4: No AB, no X, ${6\choose 4}=15$.

The sum is $45$.