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Let $G$ be a graph with $k$ vertices and $e$ edges and let $X$ be the number of copies of $G$ in $G(n,p)$. The expectation of $X$ is $$ {n \choose k}\frac{k!}{|aut(G)|} p^e = \Theta \left( n^kp^e\right) $$

I am wondering how we can obtain $\Theta \left( n^kp^e\right)$. I can see that ${n \choose k}\frac{k!}{|aut(G)|} p^e\leq \frac{n^k}{k!}k!p^e=n^kp^e$, but I could not find any lower bound for it!

Brian Moehring
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Jivid
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  • Perhaps this helps: https://math.stackexchange.com/questions/1265519/approximation-of-combination-n-choose-k-theta-left-nk-right?noredirect=1&lq=1 – Gary Jan 02 '22 at 04:12

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Hint

$$A=\binom{n}{k}\,k!=\frac {n!}{(n-k)!}$$ Take logarithms of both sides, use Stirling approximation twice and continue with Taylor series for "large" values of $n$.

When done, use Taylor again since $A=e^{\log(A)}$

Claude Leibovici
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