# Exponential Generating Function for Permutation with no Fixed Points

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

$$\sum_k\binom{n}kd_{n-k}=n!\;.$$

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

$$e^x=\sum_n(1)\frac{x^n}{n!}$$

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

$$e^xg(x)=\sum_n\left(\sum_k\binom{n}k(1)d_{n-k}\right)\frac{x^n}{n!}=\sum_n(n!)\frac{x^n}{n!}=\frac1{1-x}\;,$$

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

• A comment explaining the downvote would be helpful. Nov 4, 2014 at 3:34
• Wasn't me! Thank you very much for the help-- I thought the problem might have been much more straightforward than it was! Nov 4, 2014 at 3:37
• @user2899162: You’re very welcome. Nov 4, 2014 at 3:37
• (+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. Nov 4, 2014 at 20:11