Dice rolling question (multinomial) Three $6$-sided fair dice are rolled. In $10$ independent throws, Let $A$ be the number of times all the sides are the same and let $B$ be the number of times only two sides are the same. Find $E(6AB)$:
Here is how I approached this:
I know that $E(6AB) = 6E(AB)$. I know that $E(AB)$ for a multinomial distribution is:
$$n(n-1)\cdot\text{probability of }A\text{ occurring}\cdot\text{probability of }B\text{ occurring}$$
Probability of $A$ occurring is $6\left(\dfrac16\right)^3$. Probability of $B$ occurring is $6\left(\dfrac16\right)^2 \left(\dfrac56\right)$
Therefore $$E(AB) = 10 \cdot 9 \cdot 6 \left(\frac16\right)^3 \cdot 6 \left(\frac16\right)^2 \cdot \frac56$$ and $E(6AB)$ is just $6$ times the quantity to the left.
I would like to know if my approach above is correct since I don't get the correct solution of $25/4$ Thanks for the help.

Let $C$ be the number of throws when all dice have different outcomes. Then $A+B+C=10$, and the triple $\{A,B,C\}$ follows multinomial distribution with parameters $n=10$ and probabilities $p_1 = \frac{6}{6^3} = \frac{1}{36}$, $p_2 = \frac{3 \cdot 6 \cdot 5}{6^3} = \frac{5}{12}$ and $p_3 = \frac{6 \cdot 5 \cdot 4}{6^3} = \frac{5}{9}$. It is easy to see that $p_1 + p_2 + p_3 = 1$.
The moment-generating function for multinomial distribution is $\phi(t_1,t_2,t_3) = \left( p_1 \mathrm{e}^{t_1} + p_2 \mathrm{e}^{t_2}  + p_3 \mathrm{e}^{t_3}  \right)^{10}$.
Then $\mathrm{E}(A B) = \left. \partial_{t_1} \partial_{t_2} \phi(t_1,t_2, t_3)\right\vert_{t_i=0} = 10 \cdot 9 \cdot p_1 p_2 \cdot \mathrm{e}^{t_1+t_2} \cdot \left( p_1 \mathrm{e}^{t_1} + p_2 \mathrm{e}^{t_2}  + p_3 \mathrm{e}^{t_3}  \right)^7\vert_{t_i=0} = 90 p_1 p_2 $. 
Therefore $\mathrm{E}(6 A B) = 6 \mathrm{E}(A B) = 6 \cdot 90 \cdot p_1 p_2 = \frac{25}{4}$.
