Since there are n even numbers in $[2n]$ we know we will have $n$ cycles.
If length of all cycles is 2 we have $n!$ ways to make a permutation which satisfies given condition.
We can also have some cycles of length 3(they contain one even and two odd numbers) and for each of those we have a fixed point and it must be even.
Its easy to notice that the highest number of pairs of cycles of lengths 1 and 3 is whole part of the number $\frac{n}{2}$ (I don't know how to write whole part and this sentence is generally terrible but bear with me)
So now I want to see what happens when we have $k$ pairs of cycles of length 1 and 3.
We choose two even and two odd numbers in $n \choose 2$$n \choose 2$ ways.Now we can make 4 permutations out of those numbers(we choose cycle of length 1 in two ways because it must be even and then we choose a cycle of length 3 out of numbers that are left)
We repeat this $k$ times (for last pair of cycles we choose two even and two odd numbers in $n-2(k-1) \choose 2$$n-2(k-1) \choose 2$ ways and make a permutation in 4 ways)
Now we have $2(n-2k)$ elements left ($(n-2k)$ even and $(n-2k)$ add).
Now we pair them even with odd in $((n-2k)!)^2$ ways.
Ok, so now I want to divide all of this with something(since order is not important) but I cant figure out with what.
Notice I didn't divide with anything when I was counting pairs of cycles of length 1 and 3 and when I was counting transpositions.I thought maybe I could divide it all by $n!$ at the end(now that is) and then multiply it by 2 since I was considering order when I was counting number of permutations for those pairs of cycles, but that just doesn't seem right.
I would appreciate some help with finishing this.Also alternative solution would also be greatly appreciated since I have a feeling I'm doing this the worst possible way.
Best Answer
Consider any listing of the numbers $1, 2,...,2n$ that obey the following two conditions: (1) The even numbers appear in order (not necessarily consecutively), and the last number in the list is $2n$. For example if $n=4$, the following would be a permissible list: $5,1,2,4,3,7,6,8$.
Each such listing corresponds to a permissible permutation of $1, 2,...,2n$, where a cycle ends with an even number. For instance in the example above, the permutation would be $(5,1,2)(4)(3,7,6)(8)$.
And this correspondence is reversible. Given a permissible permutation, we can get such a listing, by just putting the even number in each cycle at the end of the cycle, and then removing the cycle parentheses to get the listing.
Now start with the even numbers in their proper order $2,4,6,...,2n$. Next place $1$. There are $n$ choices for this--$1$ can go before any of the even numbers. Next place $3$. There are $n+1$ choices here ($3$ could go before any number currently in the list, which now includes $1$ and all the evens). Next place $5$ ($n+2$ ways to do this). Etc.
Altogether there are $n\cdot (n+1)\cdot (n+2)\cdot \ldots \cdot (2n-1)$ ways to construct these sequences; and consequently that is the number of admissible permutations as well.