Random Permutation Statistics - Derangements Containing An Even and An Odd Number of Cycles

Derangements Containing An Even and An Odd Number of Cycles

We can use the same construction as in the previous section to compute the number of derangements containing an even number of cycles and the number containing an odd number of cycles. To do this we need to mark all cycles and subtract fixed points, giving

$g(z, u) = \exp\left( - u z + u \log \frac{1}{1-z} \right) = \exp(-uz) \left( \frac{1}{1-z} \right)^u.$

Now some very basic reasoning shows that the EGF of is given by

$q(z) = \frac{1}{2} \times g(z, -1) + \frac{1}{2} \times g(z, 1) = \frac{1}{2} \exp(-z) \frac{1}{1-z} +\frac{1}{2} \exp(z) (1-z).$

We thus have

$D_0(n) = n! q(z) = \frac{1}{2} n! \sum_{k=0}^n \frac{(-1)^k}{k!} + \frac{1}{2} n! \frac{1}{n!} - \frac{1}{2} n! \frac{1}{(n-1)!}$

which is

$\frac{1}{2} n! \sum_{k=0}^n \frac{(-1)^k}{k!} + \frac{1}{2} (1-n) \sim \frac{1}{2e} n! + \frac{1}{2} (1-n).$

Subtracting from, we find

The difference of these two ( and ) is

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