Roll a die probability question 
An unbiased six-sided die is to be rolled five times. Suppose all these trials are independent. Let $E_1$ be the number of times the die shows a 1, 2 or 3. Let $E_2$ be the number of times the die shows a 4 or a 5. Find $P(E_1 = 2, E_2 = 1)$.

I have tried to solve this question this way:
Total $6P6 = 720$.
1,2 and 3 can be placed in $2$ locations $= 3^2$
4 and 5 can be placed in 1 location $= 2^1$
and 6 can be placed in two locations $= 2^2$
The probability is $= 72/720 = 0.1$
Is this correct?

The probability equals:$$\binom5{2,1,2}\left(\frac36\right)^2\left(\frac26\right)^1\left(\frac16\right)^2=\frac5{72}$$(where $\binom5{2,1,2}:=\frac{5!}{2!1!2!}$)
You can look at this as a case of trinomial distribution. 
There are $5$ independent experiments and there are not $2$ possible and mutually exclusive outcomes (success and failure, as is the case by binomial distribution) but there are $3$ possible and mutually exclusive outcomes: the die shows a face in $\{1,2,3\}$, or in $\{4,5\}$ or in $\{6\}$. This with probabilities $\frac36$, $\frac26$ and $\frac16$ that sum up to $1$.
