You throw a fair die $n$ times. Let $S_1$ be the number of tosses where a $1$ appears, and $S_2$ the number of tosses where a $2$ appears. Determine $\operatorname{Cov}(S_1,S_2)$.
My attempt
I started by noting that $\operatorname{Cov}(S_1, S_2)=E(S_1S_2)-E(S_1)E(S_2)$ and that since we know that $S_1$ and $S_2$ are $\sim \operatorname{Bin}(n,1/6)$, we get that $E(S_1)=E(S_2)=\frac{n}{6}$. Now, I tried calculating $E(S_1S_2)$ like this:$$E(S_1S_2)=\sum \limits_{\omega\in \{1,2,...,n\}}(S_1\cdot S_2)(\omega)P(\{\omega\}) = \sum \limits_{k=1}^{n}k^2P(X=k)=\sum \limits_{k=1}^{n}k^2\begin{pmatrix} n\\ k \end{pmatrix}\cdot \left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{n-k}$$ But I can't evaluate this sum without computer assistance. Is there a more elegant way or is mine even wrong?
You should use the linearity of the expectation and the fact that $$S_1S_2=\sum_{i,j=1}^n B_i C_j$$ where $B_i$ and $C_j$ are the Bernoulli variables such that $$S_1=\sum_{i=1}^n B_i \hspace{15pt} S_2=\sum_{j=1}^n C_j$$ If $i \neq j$ then $B_i$ and $C_j$ are independent, so $$E(B_iC_j)=E(B_i)E(C_j)=\frac{1}{36}$$ while $B_iC_i=0$. We conclude that $$E(S_1S_2)=\sum_{i,j=1}^n E(B_i C_j)=\frac{n(n-1)}{36}$$