Struggling with Combinatorics Context: I'm struggling with permutations and combinations. I find that I can be believe that I'm solid with my reasoning, however, when I check the numerical answer I'm often wrong. I might even use another reason to obtain the correct answer, however when I apply that same reasoning to another part it will get the wrong answer.
Example: 


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*Four players are chosen to compete in 2 games of 2 people. What is the probability that two particular players will be in the same team (say Jerry and Joe). I went, there is only one way to arrange Joe and Jerry such that they're in the same team. However, I can pick the 2 teams of 2 in $ \frac {\binom{4}{2} \binom{2}{2}}{2!}$ (the numerator is the choosing the 2 teams of 2, and the denominator is because AB vs CD is the same as CD vs  AB). Finally I arrive at the result; $P(E)=\frac{1}{3}$ (this was the correct answer).

*The next question was similar but, it 8 players instead of 4, but still 2 games of 2 people and probability of two particular players in same team. Ok, so from before reasoning, the number of ways together is 1, but the number of ways to choose the team is $ \frac {\binom{8}{2} \binom{6}{2} \binom{4}{2} }{4!}$. Thus $P(E)=\frac{1}{105}$. However this is incorrect. So I reasoned, maybe the number of ways the two particular people is the number of ways in total (105) subtract the number of ways not together. Ok, so I'll pick the Jerry to be in the first team, thus for Joe to not be with him he must go in the other 3 teams, BUT I must then place the other 6 people into any team ($6!$), but divide by $4!$ as the teams are interchangeable. So then $P(E)=\frac{105 - \frac{3\times6!}{4!}}{105} = \frac{1}{7}$ (the correct answer).


Though, by applying the same reasoning to the first example I don't get the correct answer (I will not list it here for brevity).
Question: How can I be certain that my reasoning is correct for a question? And how can I overcome this barrier with combinatorics?
In calculus, if I was asked to find the derivative at a point, or what not, I could clearly either see the flaw in my logic (divide by zero, didn't consider the range of a function, plug it into a calculator etc.), however I find that with combinatorics I'm grasping at straws, hoping my solution is correct.
Thanks
 A: One useful technique for counting problems such as this is to label the objects (in this case, players) and then counting the ordered outcomes. Imagine each player has an ID: Player 1 all the way to Player 8, and let's say we want Players 1 and 2 (Jerry and Joe) to always be on the same team.
In the first case with 4 players, we can represent each outcome as a string of digits constructed using the ID numbers of each player. For instance, we can take 1234 to mean that Players 1 and 2 are on one team, while Players 3 and 4 are on the other. There are 4! ways to arrange this string of four digits. As for the favorable outcomes: since 1234 is essentially the same as 2143, we have 2! 2! such permutations (orderings within each team), and then we multiply by another 2! because 1234 is the same as 3412 (the two teams are not distinct and we need to count the ordering of the teams themselves). Therefore, the answer for the first case is (2!2!2!)/4! = 1/3.
In the second case, we can still apply the same labeling, and we know that here, there are 8! ways to arrange the string of 8 digits. Let's say we politely ask the 8 people to queue, and decide that the first 4 spots would be the players (1st and 2nd spot as one team, 3rd and 4th spot as another team), and then those at the last 4 spots of the line as non-playing. Since we want Players 1 and 2 to always be included as players, suppose we first fix them in the first 2 spots of our queue. Then there are 6! ways to get the other 6 people to queue. Once we have all these possibilities, we again multiply by the same 2! 2! 2! as in the 4-person case to account for the same possibilities (orderings within each team and the orderings of the two teams themselves). Therefore, the answer in the 8-person case is (6!2!2!2!)/8! = 1/7.
