Number of Permutations of a Cycle with no cycle greater than 11. I'm looking for the number of permutations of a $20$ element set, with no cycle greater than length $11$. All of the attempts that I have tried are not working out. I'm not sure how to get started on this question. Any hints?
Also, I saw a similar question to this using generating functions. But I'm doing it without, if possible.
 A: We count the complement, the permutations that have a cycle of length $k$, where $k$ ranges from $12$ to $20$. Then we add up. For smallish $k$, like $5$, this could be unpleasant, since one can have several $5$-cycles. But for the $k$ in our range, a permutation has at most one $k$-cycle. 
Let's do the calculation. The idea has already been covered by amWhy, so we do it quickly.
The $k$ objects in the $k$-cycle can be chosen in $\binom{20}{k}$ ways. There are $(k-1)!$ circular permutations of $k$ objects. And there are $(20-k)!$ ways to permute the rest, for a total of 
$$\binom{20}{k}(k-1)!(20-k)!.$$
This simplifies nicely to $\dfrac{20!}{k}$. So the complement of our set has size
$$20!\left(\frac{1}{12}+\frac{1}{13}+\cdots+\frac{1}{20}\right).$$
Remark: The result may be numerically surprising. The probability that a permutation has a large cycle is quite big. 
A: Whenever a permutation contains any cycle of length 11, that one has to be the longest cycle in the permutation, because there are not enough elements left over to form a longer one. So what we're looking for is simply all permutations containing an 11-cycle. These all consist of 


*

*a cycle of length 11, 

*some permutation of the remaining 9 elements.


There are $\,\displaystyle \binom{20}{11}\cdot 10!\,$ different $11$-cycles, because we can first choose which $11$ elements the cycle includes. For each such choice there are $11!$ different cycle notations, but every cycle is counted 11 times (with each of the 11 elements notated first), so there are actually $10!$ different cyclic permutations of 11 elements.
Once an $11$-cycle has been chosen, then there are $9!$ ways to permute the remaining $9$ elements, so the total number of permutations that contain an $11$-cycle is given by:
$$ \binom{20}{11} \cdot 10! \cdot 9! = \frac{20!}{11}$$
If you also need to count permutations whose longest cycle is LESS THAN 11, then there is much more work involved, and indeed, I believe @André Nicolas's answer involves the least amount of work.
