"How many ways can you partition a set of size $n$ into $k$ parts of the same size?"
I've tried solving in the following way but I'm not sure if it's correct, feedback would be appreciated. I'm new to combinatorics.
If $k$ does not divide $n$ the answer is trivially $0$. Otherwise, each part must have $\frac{n}{k}$ elements. The first of the $k$ parts can be chosen to be any $\frac{n}{k}$-subset of the $n$-set, and there are $\binom{n}{n/k}$ such subsets. The second part is a $\frac{n}{k}$-subset of the remaining $n-\frac{n}{k}$ elements, so there are $\binom{n-\frac{n}{k}}{n/k}$ subsets to choose from. Continuing in this fashion, the total number of ways to select parts becomes:
$\binom{n}{n/k}\times\binom{n-\frac{n}{k}}{n/k}\times\binom{n-2\frac{n}{k}}{n/k}\times\ldots\times\binom{{2n/k}}{n/k}$
However, since the order in which the subsets are selected doesn't matter, I have to divide by the number of permutations of $k$ parts, which is $k!$ Hence the total number of partitions becomes:
$\frac{\binom{n}{n/k}\times\binom{n-\frac{n}{k}}{n/k}\times\binom{n-2\frac{n}{k}}{n/k}\times\ldots\times\binom{{2n/k}}{n/k}}{k!}$
Is there anything wrong with this solution?