Calculate the probability in winning tennis game. 
A tennis tournament has $2n$ participants, $n$ Swedes and $n$ Norwegians. First, $n$ people are
  chosen at random from the $2n$ (with no regard to nationality) and then paired randomly with
  the other $n$ people. Each pair proceeds to play one match. An outcome is a set of $n$ (ordered)
  pairs, giving the winner and the loser in each of the $n$ matches. 
(a) Determine the number of
  outcomes. 
(b) What do you need to assume to conclude that all outcomes are equally likely?
(c) Under this assumption, compute the probability that all Swedes are the winners.

My work:
So out of $2n$ people we can choose $n$ people in $2n \choose n$ ways. And then pair them together, so we get $2n\choose n $$n!$ ways total. Now I am unable to calculate the number of outcomes for the first question.
For (b), I assume that every player is equally able i.e equal strength and all.
For (c), we need to pair Norwegians against Swedes, so that would be $n!$ ways.
So help me to solve (a), and check my solutions for (b) and (c). Thanks.
 A: a) First choose the $n$ people who will be the winners. Line them up in order of height. For each choice of winners, there are $n!$ ways to choose the people they beat, for a total of $\binom{2n}{n}n!$ outcomes.
Your b) and c) are right. The number of pairings is not $\binom{2n}{n}n!$. For the number of pairings we would have to divide that by $2^n$.
A: Probability can get confusing so my strategy has always been to break the problem into discrete chunks. Consider part (a) and the small questions we should consider. 
Breaking the problem down


*

*There are 2$n$ participants. How do we choose $n$ people to form group 1 and let the rest form group 2? We'll denote the groups $G1,G2$ respectively.

*Given we have $G1,G2$. It's fixed and constant (these groups don't change). How do we consider all the possible pairs $(g1,g2)$ where $g1 \in G1$ and $g2 \in G2$? 

*If each pair $(g1,g2)$ plays a match. How many outcomes can there be?


Some thoughts and hints
Now that we've broken the problem down, the problem seems at least a little more tractable. Let's see if we can systematically go through those steps. 
Your intuition is actually mostly correct, so let me help just by guiding you along. 


*

*There are exactly $\binom{2n}{n}$ ways to choose $n$ people to form $G1$. The rest form $G2$. This seems straightforward.

*Now that our groups are fixed. Let's say we'll pick one of $n$ people in $G1$ and one of $n$ people in $G2$ to form our first pair. To form the second group, we can do the same and pick one of the remaining $n-1$ people from the first group and one of the remaining $n-1$ people from the second group. 


This comes out to be
$$ \text{total combinations} = \binom{n}{1} \binom{n}{1} \times \binom{n-1}{1}\binom{n-1}{1} \cdots \binom{1}{1}\binom{1}{1} $$ 
But we have to be careful here because we have counted repeats. Why? How do we account for it? Seems like you got it. There are $n!$ ways to order pairs. In other words, we don't want to get the same pairings but in different orders. So this part comes out to be 
$$ \text{total combinations} = \frac{\binom{n}{1} \binom{n}{1} \times \binom{n-1}{1}\binom{n-1}{1} \cdots \binom{1}{1}\binom{1}{1}}{n!} = n!$$


*Since we solved parts 1 and 2, we know the all possible games that can appear. But for the time being, just think about 1 game. How many outcomes are there? There are exactly 2 outcomes, one player either wins or loses. 


Notice that part 1, part 2, part 3 are all independent from each other. Therefore, the total number of outcomes is
$$ \text{total outcomes} = \binom{2n}{n} \times n! \times 2 $$
What we've essentially done is compute the sample space, which is just a fancy way of saying we've computed all possible outcomes. Now in part 3, you're asked to calculate the probability that only Swedes win. Think about what needs to happen for that outcome to happen. Intuitively, you know that every single one of your matches has to be swede vs. norwegian. How is that different from your original problem formulation? Looks like you got the right idea. 
