While reviewing, I've come across a problem that seems to outline my lack of knowledge with regards to (specifically exponential) generating functions. For some reason, I understand "ordinary" generating functions fine in most contexts, but I cannot grasp exponential generating functions nearly as well.

The problem is as simply stated in the title:

Find the exponential generating function for the count of permutations with no Fixed Points.

The book states the answer (might be?), $\huge{\frac{e^{-x}}{1-x}}$, but doesn't give an explanation. Of course, knowing what the solution looks like and substituting the infinite series for $\frac{1}{1-x}$, I can see that the answer is $(1+x+x^2+x^3+...)(1-x+\frac{x^2}{2} -...)$. I can't even justify this generating function, let alone derive it on my own.

I've tried splitting the problem into smaller sub-problems (first take an element in a random permutation with no fixed points of length $n$, it has chance $\frac{1}{n-1}$ of being in a cycle of a length $\{2,3,...,n\}$...) but this clearly goes nowhere as it's much more of a probabilistic argument, and not really a "combinatoric" argument that I'm trying to understand.


Let $d_n$ be the number of derangements of $[n]$ (i.e., the number of permutations of $[n]$ with no fixed points). The number of permutations with $k$ fixed points is $\binom{n}kd_{n-k}$: there are $\binom{n}k$ ways to choose the $k$ fixed points, and $d_{n-k}$ derangements of the other $n-k$ points. There are $n!$ permutations altogether, so


Now take the exponential generating function on the lefthand side. To do this, let $$g(x)=\sum_nd_n\frac{x^n}{n!}\;,$$ the exponential generating function for the derangement numbers. We also know that


is the exponential generating function for the sequence $\langle 1,1,1,\ldots\rangle$. Now take the convolution of these two exponential generating functions:


so $$g(x)=\frac{e^{-x}}{1-x}\;.$$

  • $\begingroup$ A comment explaining the downvote would be helpful. $\endgroup$ – Brian M. Scott Nov 4 '14 at 3:34
  • $\begingroup$ Wasn't me! Thank you very much for the help-- I thought the problem might have been much more straightforward than it was! $\endgroup$ – druckermanly Nov 4 '14 at 3:37
  • $\begingroup$ @user2899162: You’re very welcome. $\endgroup$ – Brian M. Scott Nov 4 '14 at 3:37
  • $\begingroup$ (+1) Very nice work. By way of enrichment this MSE link I may prove useful reading, as might this MSE link II and this MSE link III. $\endgroup$ – Marko Riedel Nov 4 '14 at 20:11

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