Re: Symmetric group counting

There are $\displaystyle \binom{n}{2} = \dfrac{n(n-1)}{2}$ ways to choose two elements for the first 2-cycle. There are $\displaystyle \binom{n-2}{2} = \dfrac{(n-2)(n-3)}{2}$ ways to choose two elements for the second 2-cycle. Since you can choose the 2-cycles in either order, this double counts:

$\displaystyle \dfrac{\binom{n}{2}\binom{n-2}{2}}{2} = \dfrac{n(n-1)(n-2)(n-3)}{8}$

Re: Symmetric group counting

Quote:

Originally Posted by

**SlipEternal** There are $\displaystyle \binom{n}{2} = \dfrac{n(n-1)}{2}$ ways to choose two elements for the first 2-cycle. There are $\displaystyle \binom{n-2}{2} = \dfrac{(n-2)(n-3)}{2}$ ways to choose two elements for the second 2-cycle. Since you can choose the 2-cycles in either order, this double counts:

$\displaystyle \dfrac{\binom{n}{2}\binom{n-2}{2}}{2} = \dfrac{n(n-1)(n-2)(n-3)}{8}$

Thanks. I had thought that there might be a combinatorial in there somewhere. I just couldn't find it.

-Dan