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I attempted this on my own and got a fairly simple solution. However, after reading proofs here and here, I feel like I have massively over simplified the problem. I understand the other solutions, but does my shorter one work?

Proposition. The expected number of cycles of length $k$ in an Erdős–Rényi random graph $G_{n,p}$ is $$\binom nk(k-1)!p^k/2.$$

Proof. A cycle of length $k$ requires $k$ vertices of which there are $\binom nk$ choices. There are $k!$ different cycles/paths we could potentially follow in a cycle on these $k$ vertices. Since the starting point is irrelivant, we can divide by $k$, as is the direction/orientation, giving $(k-1)!/2$ different cycles. For each cycle, we require the $k$ edges between them to be included in the graph, which happens with probability $p^k$. Conclude that the expected number of such cycles is as stated.

Edit: I am only interested in cycles of length $k$, sorry!

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Szmagpie
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  • Apologies for my erroneous answer. Your solution is basically the same as this one here. – angryavian May 24 '16 at 04:20
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    @angryavian I had seen that but was worried about independence, to be honest. Was your answer incorrect because $E(Y+Z)=E(Y)+E(Z)$ for any RVs $Y$, $Z$ (i.e. independence is not necessary for the final step in my proof)? – Szmagpie May 24 '16 at 04:29
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    Yes, basically. That is the advantage of using indicators. – angryavian May 24 '16 at 04:30
  • The solution above makes sense to me, but I'd like to ask some further questions, since I usually get confused when analyzing similar situations. In particular, why is it not enough to just stop at ${n \choose k}$ (and multiplying it by the probability term of course), but we have to go on with the analysis of $k!$ ways the elements are arranged? I don't see the train of thoughts here. – ensbana Feb 15 '20 at 14:41

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