this is how i count ncycles in Sn (this works for any n, but i will use n = 4):
since any 4cycle in S4 involves 1, we may as well start with it:
(1.....
i have 3 choices for my next element (the image of 1):
(1 2.....
(1 3.....
(1 4.....
after i make my next choice, i'll have them all:
(1 2 3 4)
(1 2 4 3)
(1 3 2 4)
(1 2 4 2)
(1 4 3 2)
(1 4 2 3)
that makes 6.
in general, in Sn, we'll get (n1)(n2)....(2) = (n1)! possible distinct ncycles.
