How to count permutations with cycles of length at least 51 in $S_{100}$? Let consider permutation $ \in S_{100} $ How to count the number of permutations of those which contains a cycle of length 51 at least. ( so I would like a cycle of length 52,53,54,....,100)
 A: I am not sure if this is the easiest way or the type of answer you are looking for but one thing to notice is that since 51>100/2=50, we know that any permutation with a cycle of length 51, the only other cycles it can contain are less than 50. So for example if $c(51)$ is the number of cycles with length 51 and $p(49)$ are the number of permutations on $49$ elements then the number of of permutations in $S_{100}$ with length $51$ cycles is $c(51)p(49)$ since there is no double counting due to the fact a permutation on $49$ elements can not contain a permutation of length $51$.
A: For each $k>50$, we claim there are $100!/k$ permutations with a $k$-cycle.  To see this, write down the numbers from $1$ to $100$ in arbitrary order; declare the first $k$ to be a cycle, and interpret the remaining $100-k$ symbols as a permutation.  As there are $k$ rotations of the cycle, the claim follows.  Summing from $k=51$ to $100$, the final sum is
$$100!\ (H_{100}-H_{51})$$
which agrees with Marko Riedel's answer.
