A fair die is rolled 14 times. Let X be the number of faces that appear exactly 2 times. Find Var(X). To start, we know $\mathsf{Var}(X) = \mathsf{E}(X^2) - \mathsf{E}(X)^2$. 
I calculated that the $\mathsf{E}(X)$ is $6\times C(14, 2)(1/6)^2(5/6)^{12}$
How do I find $\mathsf{E}(X^2)$? 
I'd really appreciate the answer since I have an exam tomorrow and this type of question might come up. 
I found $\mathsf{E}(X)$ using the method of indicators. 
Thank you for any help!
 A: Your efforts will probably go along with what I write here.
For $i=1,\dots,6$ let $X_i$ take value $1$ if face $i$ appears exactly twice, and takes value $0$ otherwise.
Then: $$X=\sum_{i=1}^{6}X_{i}$$ so that:$$\begin{aligned}\mathsf{Var}X & =\mathsf{Covar}\left(\sum_{i=1}^{6}X_{i},\sum_{i=1}^{6}X_{i}\right)\\
 & =\sum_{i=1}^{6}\sum_{j=1}^{6}\mathsf{Covar}\left(X_{i},X_{j}\right)\\
 & =6\mathsf{Var}X_{1}+30\mathsf{Covar}\left(X_{1},X_{2}\right)
\end{aligned}$$where the second equality rests on the bilinearity of $\mathsf{Cov}$ and the third equality on symmetry.
Here $X_1$ has Bernoulli distribution with parameter $p=\binom{14}2\left(\frac16\right)^2\left(\frac56\right)^{12}$.
(this agrees with what you found yourself)
Also $X_1X_2$ has Bernoulli distribution with parameter $p'=\frac{14!}{2!2!10!}\left(\frac16\right)^4\left(\frac46\right)^{10}$.
(do you have troubles to understand this?)
This allows you to find $\mathsf{Var}X_1$ and $\mathsf{Covar}\left(X_{1},X_{2}\right)$.
A: This looks like self-study so I will give some hints. Define first 6 indicator random variables $Y_1, i=1,,2,3,4,5,6$ counting the number of times $i$ eyes show. $Y$ have a symmetric multinomial distribution, $Y \sim \text{multinom}(\frac16, \dotsc, \frac16, n=14)$. Then define $I_i$ the indicator on the event $Y_i=2$. Calculate the covariance matrix of the vector random variable $Y$ and use the well known formula $\DeclareMathOperator{\V}{\mathbb{V}}$  $\V a^tY = a^T \V(Y) a$.
