I want to count all 2-cycle permutations of a set S = {1,2,3,4,..,n}.
After some calculations I came to following result: $ \sum_{k=1}^{k=n-1} {n \choose n-k} * k!* (n-k)!$
I got this result by detection, that if n equals 4, then permutations are:
(123)(4) --> I have to choose 3 numbers into first permutation and permutate them and permutate the remaining one number.
(12)(34) --> I have to choose 2 numbers into first permutation and permutate them and permutate the remaining 2 numbers.
(1)(234) --> I have to choose 1 number into first permutation and permutate it and permutate the the remaining 3 numbers.
Is this the right approach or I forgot some more permutations?
I think you were on the right track. However, for a $k$-cycle, there are $k$ different ways to write it. So dividing by $k$ will be necessary.
Also, your original formula double counts. For, disjoint cycles commute.
Adjusting, we get $\sum_{k=1}^{\lfloor n/2\rfloor}{n\choose k}(k-1)!(n-k-1)!$.