Quote Originally Posted by HallsofIvy View Post
A product of even numbers can be reduced as:
2(4)(6)\cdot\cdot\cdot(2n-2)(2n)= [(1)(2)][(2)(2)][(3)(2)]\cdot\cdot\cdot[(n-1)(2)][(n)(2)] = n!2^n.

A product of odd numbers is more complicated- you have to put the even numbers in:
(1)(3)(5)\cdot\cdot\cdot (2n-1)(2n+1)= \frac{(1)(2)(3)(4)(5)\cdot\cdot\cdot(2n-1)(2n)(2n+1)}{2(4)(6)\cdot\cdot\cdot(2n-2)(2n)}= \frac{(2n+1)!}{2^n n!}.

For the record: I know this.