Problem: Let $I=\{1,2,\ldots,2n\}$ be a set which is partitioned into $n$ subsets $X_i$, each containing exactly $2$ elements such $\bigcup\limits_{i=1}^n X_i=I$. Prove that the number of ways this can be done is $\dfrac{(2n)!}{2^n\cdot n!}$
Attempted solution:
First, let us consider the problem of partitioning $I$ into $n$ 2-tuples such that every element of $I$ goes into exactly one tuple. We count the number of ways of doing this.
First, we can select the first element of the $n$ 2-tuples by selecting $n$ elements from $I$ (containing $2n$ elements). This can be done in $\dbinom{2n}n$ ways. Now, the second element of the $n$ 2-tuples are taken from the rest of the (already determined) $n$ elements of $I$. This can be done in $n!$ ways, so by the rule of product, there are $\dbinom{2n}n n!$ ways.
Now, consider relaxing the 2-tuples (ordered) to sets with 2 elements (unordered). Since there are $2$ elements, we count each set $2!$ times and by the rule of product, for a configuration of $n$ such sets, we count each configuration $(2!)^n=2^n$ times.
Hence, the number of ways for the original problem is $\dfrac{\dbinom{2n}n n!}{2^n}=\dfrac{(2n)!}{2^n\cdot n!}$
Is my thinking correct?
If yes, can't this argument be generalized for $I=\{1,2,\ldots,kn\}$ where $k$ is a positive integer and we're partitioning $I$ into $n$ subsets $X_i$, each containing exactly $k$ elements?
By an argument analogous to the one above, the number of ways seems to be,
$$\frac{\binom{kn}n\binom{(k-1)n}n\cdots\binom{2n}n n!}{(k!)^n}=\frac{n!\prod\limits_{m=2}^k\binom{mn}n}{(k!)^n}$$
Is my thinking and conclusions correct?