I did a proof a few days ago (for the sake of enjoyment) and my teacher thought it was interesting, though he seemed unsure of my result.(adsbygoogle = window.adsbygoogle || []).push({});

Consider a set of n distinct objects, P. If [itex]n \in \mathbb{Z}_+ \cup \left\{0\right\}[/itex], then the cardinality, q, of the set of all derangements of P is given by

[itex]\displaystyle q = \frac{n!}{2\pi i}\oint_{\gamma}\Gamma(z)dz[/itex],where [itex]z=a+bi[/itex] and [itex]\gamma[/itex] is given by the parametric equations [itex]a = (\frac{n}{2}+\frac{1}{\alpha})cos(t)-\frac{n}{2}[/itex] and [itex]b = (\frac{n}{2}+\frac{1}{\alpha})sin(t)[/itex] for some arbitrary [itex]\alpha > 1[/itex]. Note that the case of n=0 is almost by convention, much like 0!=1.

My proof used the transitive property of equality (the formula for derangements and for the sum of the residues of the gamma function are the same). Is there a more fundamental reason why they happen to be the same?

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# Derangements and Contour Integrals?

Can you offer guidance or do you also need help?

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