If you have $2n$ people how many different ways can you pair them up? If you have $2n$ people how many different ways can you pair them up?
My method:
Considering making pairs like this:
Randomly line them up and then pair of the $1^{st}$ person in line with the $2^{nd}$, the $3^{rd}$ with the $4^{th}$, ..., the $(2n-1)^{th}$ with the $2n^{th}$
We have $2n!$ ways to line them up. However, this overcounts pairs, a lot.
Consider just 6 people. $A,B,C,D,E$ and $F$. The line $ABCDEF$ gives the same result as the line $BACDEF$
We can flip each person with their partner. He can do this $n$ times so we have overcounted $2^n$ times per order.
However, we have still overcounted! Consider again the line $ABCDEF$ this gives the same pairs as $ABEFCD$ We can indeed shuffle the pairs , as long as they remain with their partners in any order! We have $n!$ ways to shuffle the pairs.
My final answer is then $\frac{2n!}{2^n n!}$
Is this correct? Is anyone able to give some pointers as to how to use some code to confirm this?  Does someone have a nicer answer?
 A: Your answer is correct, but you should write it as $\frac{(2n)!}{2^n n!}$.
Other possible solution is the following. Take any person. Choose a couple for that person, that can be done in $2n-1$ ways. Take any of the remaining $2n-2$. Choose a couple for that person, that can be done in $2n-3$ ways. Continue this process until all couples are assigned. The total number of choices is $$(2n-1)(2n-3)\cdots 3\cdot 1 = \frac{(2n)(2n-1)(2n-2)\cdots 3\cdot 2\cdot 1}{(2n)(2n-2)\cdots2}=\frac{(2n)!}{2^n n!}$$
A: Your solution is correct (make sure to write $(2n)!$ instead of $2n!$). I'll give you another method, which I think is easier to understand.
Let $T(n)$ be the number of possible pairings of $2n$ people. If we take the fist person, it can be paired with $2n-1$ different people. Once you did this you have $2n-2$ people left, which proves that $T(n)=(2n-1)T(n-1)$. Now, it is trivially to show that $T(2)=1$.
Hence we have reduced the problem to a recurrence equation, wich is easy to solve and the solution is:
$$T(n)=(2n-1)(2n-3)\dots 1=\frac{(2n)!}{2^n n!}$$
For me, and probably for many of us, transforming a combinatoric problem to a linear recurrence equation (which can be done in many of the exercises at undergad level) makes it easier, since it is more standard. In a harder problem, a solution like yours can be very difficult to find.
A: Your reasoning is perfectly fine and your answer is correct.
Another approach is to use binomial coefficients (yet very parallel to your method).

*

*How many ways are there to form the first couple? ${2n \choose 2}$.

*The second couple? ${2n-2 \choose 2}$.

*The third? ${2n-4 \choose 2}$.

*etc.

Therefore, there are
$${2n \choose 2} {2n-2 \choose 2} {2n-4 \choose 2} \dots {2 \choose 2} = \frac{(2n)!}{2^n}$$
ways to choose the couples (observe the cancellations in the product).
Finally, as you argued above, the couples may be ordered in $n!$ ways. Hence, the final answer is
$$\frac{(2n)!}{2^n n!}.$$
