Number of Unordered Partition of $[2n]$ into $n$ pairs First the notation $[2n] = \{1,2,....,2n\}$
Since partition is a set of disjoint and non-empty subset of [2n], 
I assign each element in partition one element in [2n] : $2nCn$
Now we have left n elements to assign, and each n elements has n possibilities to be assigned: $n^n$
Therefore, the answer is $2nCn * n^n$ 
However, the solution says $(2n-1)!!$.
Is the number $2nCn*n^n = (2n-1)!!$ ?
 A: You get $n^n$ by matching each of your first $n$ elements one of the $n$
remaining elements. But one cannot assign the same element to two different
ones in your first $n$. So this number should be $n!$ not $n^n$.
But is the answer $\binom{2n}n n!$? No, since this overcounts. You started
by picking an element from each pair in the matching. There are $2^n$
ways to do this, so you overcounting by a factor of $2^n$. The answer is
$2^{-n}\binom{2n}nn!$.
A: Your answer is incorrect.
Method 1:  Choose two of the $2n$ elements for the first subset, two of the $2n - 2$ remaining elements for the second subset, two of the $2n - 4$ remaining elements for the third subset, and so forth.  That yields
$$\binom{2n}{2}\binom{2n - 2}{2}\binom{2n - 4}{2} \ldots \binom{2}{2}$$
ways to divide $2n$ elements into $n$ subsets of size $2$.  However, the order in which we select the subsets does not matter.  Hence, we must divide the above result by $n!$ to obtain
$$\frac{1}{n!}\binom{2n}{2}\binom{2n - 2}{2}\binom{2n - 4}{2} \ldots \binom{2}{2}$$
Method 2:  We list the elements of $[2n]$ from smallest to largest.  We have $2n - 1$ ways of selecting the element that will be matched with $1$.  That leaves $2n - 2$ elements.  We have $2n - 3$ ways of matching an element with the smallest element remaining.  Continuing in this way, we can divide the set $[2n]$ into $n$ subsets of size $2$ in 
$$(2n - 1)!! = (2n - 1)(2n - 3) \ldots (3)(1)$$
ways.
Observe that our answers are equivalent since 
\begin{align*}
& \frac{1}{n!}\binom{2n}{2}\binom{2n - 2}{2}\binom{2n - 4}{2} \ldots \binom{2}{2}\\
\qquad & = \frac{1}{n!} \cdot \frac{(2n)!}{2!(2n - 2)!} \cdot \frac{(2n - 2)!}{2!(2n - 4)!} \cdot \frac{(2n - 4)!}{2!(2n - 6)!}  \cdots \frac{2!}{0!2!}\\
\qquad & = \frac{1}{n!} \cdot \frac{(2n)!}{2!^n}\\
\qquad & = \frac{1}{n!} \cdot \frac{(2n)(2n - 1)(2n - 2)(2n - 3)(2n - 4) \ldots (4)(3)(2)(1)}{2^n}\\
\qquad & = \frac{1}{n!} \cdot n(2n - 1)(n - 1)(2n - 3)(n - 2) \cdots (2)(3)(1)(1)\\
\qquad & = (2n - 1)(2n - 3) \cdots (3)(1)
\end{align*}
