Cross-posting here as I didn't get a satisfactory solution on cv.
This is a well-known problem in random graph theory, where we show that if $X$ is the number of triangles in $G(V,E,p)$ with $p=\omega(\frac{1}{n})$, we can show that $$ P(X \geq 1) \geq 1-o(\frac{1}{n}) \to_n 1 $$ using Chebyshev inequality and asymptotic expansion of binomial coefficients, $\binom{n}{k} \sim \frac{n^k}{k!}$.
What I don't understand is the part where the second moment is used for indicator variables: $$ \mathbf{E}X^2 = \mathbf{E} (\sum_{k=1}^{\binom{n}{3}} X_k)^2 = \sum_{k=1}^{\binom{n}{3}} \mathbf{E}X^2_k +\sum_{k \neq j}\mathbf{E}X_k X_j $$ Specifically, I don't understand the absence of 2 in front of the second sum. The explanation is that triangles are ordered. This means that selecting triangle $X_j$ and then $X_k$ is different to $X_k$ and then $X_j$. Essentially I don't understand why $\mathbf{E}X_k X_j \neq \mathbf{E}X_j X_k$. If I solve it with 2 in front of the sum, I don't get the convergence.
The square of the sum is expanded into a sum of products. Both $\mathbf EX_1X_2$ and $\mathbf EX_2X_1$ occur in the expansion. You can either leave them like that and write $\sum_{k\ne j}$, or you can combine them and write $2\sum_{k\lt j}$, that's just a matter of convenience. What you can't do is write $2\sum_{k\ne j}$, because then you're counting each pair four times, not just twice.