Probability to form boy-girl pairs n pairs are formed from n girls and n boys randomly. What is the probability that each pair is formed by one boy and one girl?
 A: Line up the girls, say by student number, and let them choose a partner one at a time. The probability the first girl chooses a boy is $\frac{n}{2n-1}$. Given that she chooses a boy, the probability the second girl chooses a boy is $\frac{n-1}{2n-3}$. 
Given that the first $2$ girls each choose a boy, the probability the third girl chooses a boy is $\frac{n-2}{2n-5}$. And so on. Now take the product.
Remark: The answer can be made more compact. At the bottom we have $(2n-1)(2n-3)(2n-5)\cdots (1)$. This is $\frac{(2n)!}{2^n n!}$, so an alternative form of the answer is 
$\frac{2^n (n!)^2}{(2n)!}$. 
A: How many ways are there to form pairs without taking gender into account? The first person can choose a partner from $(2n-1)$ other people, the second from $(2n-3)$ since the first pair will already have been removed, and so on. So you have $1\cdot3\cdot5\cdots(2n-3)\cdot(2n-1)$ possible pairings, and each of them has equal probability. Some people write this as a double factorial $(2n-1)!!$, while others may prefer to expand this using single factorials as $\frac{(2n)!}{2^nn!}$.
And how many ways are there for boy-girl matchings? Well the first girl can pick from $n$ boys, the second from $n-1$ boys and so on, so there will be $1\cdot2\cdot3\cdots(n-1)\cdot n=n!$ such pairings.
The chances of a random pairing to be a boys-girls pairing are the number of boys-girls pairings divided by the total number of pairings, hence
$$p=\frac{n!}{(2n-1)!!}=\frac{n!}{\frac{(2n)!}{2^nn!}}=\frac{2^n(n!)^2}{(2n)!}$$
Sanity checks: for $n=1$ this gives you $p=1$ as expected. For $n=2$ this gives $\frac23$. Which makes sense: the first girl can choose either one of the two boys or the other girl, so the chances of her choosing a boy should be twice as high as the chances of her choosing a girl (at least if the choice is random). After that choice, everything is fixed.
The result above of course agrees with the one André Nicolas obtained in his answer. The main difference is that he does probability quotients first and products afterwards, while I'm doing it the other way round. Both approaches are valid, but depending on your state of mind you might find one of them more intuitive than the other.
