Let $(X_n)_{n \ge 1}$ be independent with $E(X_k) = \mu$ and $E((X_k)^4) \le C$ for some $C>0$. Prove that if $S_n = X_1 + ... + X_n$ then $\frac{S_n}{n} \to \mu$ almost surely as $n \to \infty$.
The proof begins by assuming $\mu = E(X_k) = 0$, then claims
$$E((S_n)^4) = E \left( \left( \sum_{k=1}^n X_k \right)^4 \right) = \sum_{i,j,k,l = 1}^n E(X_i X_j X_k X_l) = \sum_{i=1}^n E((X_k)^4) + 6 \sum_{1 \le i \lt j \le n} E((X_i)^2(X_j)^2)$$ and it is with this final equality that I'm having trouble.
I understand that all the terms containing odd exponents of $X_k$ are zero because of independence, but its the counting of the final term I'm struggling with. Why is the coefficient 6? My best explanation is that we have $n \times n$ possible arrangements of the $i$'s and $j$'s, but then remove the cases when $i=j$ because those are in the first term, so we have $n(n-1)$ arrangements. But we also have to consider the $4 \choose 2$ ways we can pick which indices of $i,j,k,l$ become our new $i,j$ (that doesn't really sound right either but I'm very confused). Our final sum only counts variations where $i < j$, so do we have to half our total of $6n(n-1)$ arrangements meaning there are $3n(n-1)$ values of $E((X_i)^2(X_j)^2) = E((X_i)^2)^2$ (this equality follows by independence) on the end?
I'd really appreciate some help with this counting argument.
For example, if $i=2$ and $j=5$, we have $6$ cases for $i,j,k,l$:
$2,2,5,5$
$2,5,2,5$
$2,5,5,2$
$5,2,2,5$
$5,2,5,2$
$5,5,2,2$
This is indeed $\binom 42$, considering choosing spots for $5$'s.
Apply this to all $1\leq i<j\leq n$. Then by independence, the final term becomes $$ 6\binom n2 E(X_i^2)^2 = 3n(n-1)E(X_i^2)^2. $$