# Symmetric group counting

• Nov 10th 2013, 12:07 PM
topsquark
Symmetric group counting
Quote:

Show that if \$\displaystyle n \geq 4\$ then the number of permutations in S_n which are the product of two disjoint 2-cycles is n(n - 1)(n - 2)(n - 3)/8.
The n(n - 1)(n - 2)(n - 3) portion is trivial. Obviously this over counts the number of permutations we're looking for and it is natural to find a counting scheme to divide by to get the number of distinct 2-cycles, but I am at a loss to explain how to get the 8.

Any thoughts?

Thanks.
-Dan
• Nov 10th 2013, 12:17 PM
SlipEternal
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}\$
• Nov 10th 2013, 12:29 PM
topsquark
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