proving how many permutations exist.
Show that the number of permutations of n numbers, for n ≥ 2, with two cycles is at most (n−1)!logn.
Unsure as to were to start any help would be appreciated.
The number of cycle permutations on $k$ letters is $(k-1)!$ If your cycles are of lengths $m$ and $n-m,$ you have $\binom{n}{m} (m-1)! (n-m-1)!$ possibilities, so your total is $$\sum_{m=1}^{n-1}\binom{n}{m} (m-1)! (n-m-1)! = n!\sum_{m=1}^{n-1} \frac{1}{m(n-m)}.$$ Can you finish from here?