> [!NOTE] Theorem > Let $n\in\mathbb{N}^{+}.$ Let $S_{n}$ denote the [[Symmetric Groups of Finite Degree|nth symmetric group]]. Let $k\in \mathbb{N}^{+}_{\leq n}.$ The number of elements that are [[Cyclic Permutation of n Letters|k-cycles]] is given by $\frac{n!}{k(n-k)!}.$ **Proof**: To specify a $k$-cycle in $S_{n}$, first we need a subset $S \subset \{ 1,2,\dots,n\}$ of size $k$. There are $n \choose k$ such subsets. Now fix $x_{1}\in S$ and let this be the starting point of the $k$-cycle $\sigma$ so $\sigma = (x_{1},x_{2},\dots,x_{k})$. There are $k-1$ options for the second entry, and once that is chosen $k-2$ options for the third entry and so on. So by [[Product Rule for Counting (Fundamental Counting Principle)|product rule for counting]], the total number of $k$-cycles is given by ${n\choose k} \cdot (k-1)!= \frac{n!}{(n-k)!k}$