What is the number of permutations of length 20 whose longest cycle is of length 11?
We first choose 11 from 20 in $\binom{20}{11}$ and the cyclic ordering can be done in $10!$ ways, for the remaining elements the total permutations or $\sum c(9,k) = 9! $
$\binom{20}{11} *10!*9! = \dfrac{20!}{11}$
Does this look right?
There are $\dbinom{20}{11}$ ways to choose a set of $11$ elements out of $20$ that will form the longest cycle. Once this set is chosen, there are $(11-1)!=10!$ different ways to arrange these $11$ elements in a cycle. Finally, there are $(20-11)!=9!$ ways to permute the remaining $9$ elements.
Therefore, there are $\dbinom{20}{11}\cdot10!\cdot9!$ ways.