Number of k-Cycles in Symmetric Group/Proof 1

Theorem

Let $n \in \N$ be a natural number.

Let $S_n$ denote the symmetric group on $n$ letters.

Let $k \in N$ such that $k \le n$.

The number of elements $m$ of $S_n$ which are $k$-cycles is given by:

$m = \paren {k - 1}! \dbinom n k = \dfrac {n!} {k \paren {n - k}!}$

Proof

Let $m$ be the number of elements of $S_n$ which are $k$-cycles.

From Cardinality of Set of Subsets, there are $\dfrac {n!} {k! \paren {n - k}!}$ different ways to select $k$ elements of $\set {1, 2, \ldots, n}$.

From Number of k-Cycles on Set of k Elements, each of these $\dfrac {n!} {k! \paren {n - k}!}$ sets with $k$ elements has $\paren {k - 1}!$ $k$-cycles.

It follows from Product Rule for Counting that:

\(\ds m\)

\(=\)

\(\ds \paren {k - 1}! \dfrac {n!} {k! \paren {n - k}!}\)

\(\ds \)

\(=\)

\(\ds \paren {k - 1}! \dbinom n k\)

Definition of Binomial Coefficient

\(\ds \)

\(=\)

\(\ds \dfrac {n! \paren {k - 1}! } {k! \paren {n - k}!}\)

\(\ds \)

\(=\)

\(\ds \dfrac {n!} {k \paren {n - k}!}\)

$\blacksquare$