I'm reading Analytic Combinatorics by Flajolet and Sedgewick, and I have an issue with the following argument from page 208:
The authors claim to derive $P(z, u) = e^{(u-1)z}/(1-z)$, the "BGF of permutations where $u$ marks the number of fixed points", except... that's false, right? It works for $u=0$ but when you substitute $u=1$ the result is $P(z, 1) = 1/(1-z)$, which would imply all permutations have exactly 1 fixed point.
Did I read this section incorrectly? The argument makes sense I think, so I don't understand why it is wrong. Does anyone know how to correct the given argument to derive the true form of $P(z, u)$?
I guess you're misinterpreting the result of substituting different values of $u$ into the BGF.
Here BGF is $$ P(z,u) = \sum_{n,k} \frac{1}{n!} P_{n,k} z^n u^k, $$ where $P_{n,k}$ is the number of permutations of $n$ elements with $k$ fixed points.
Substituting $u = 0$ gives $$ P(z,0) = \sum_{n} \frac{1}{n!} P_{n,0}z^n, $$ the EGF of permutations with $0$ fixed points (derangements).
However, substituting $u=1$ gives $$ P(z,1) = \sum_{n,k} \frac{1}{n!} P_{n,k} z^n = \sum_{n} \frac{1}{n!} \bigg(\sum_{k} P_{n,k}\bigg) z^n = \sum_n z^n, $$ the EGF of all permutations, not those with $1$ fixed point.
Similarly, substituting $u=k$ will not give an EGF of permutations with $k$ fixed points. The latter is in fact equal to $$ \frac{1}{k!}\frac{\partial^k}{\partial u^k} P(z,0). $$