1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ I think you should comment that as $9<11$ you don't get any other cycle of order $11$, and so you haven't over-counted. $\endgroup$ – ancientmathematician Oct 29 '18 at 16:25
  • $\begingroup$ @ancientmathematician right $\endgroup$ – thetraveller Oct 29 '18 at 16:38
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.