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?
 A: Close. You are rolling 5 dice, not 6. And you are not looking to permute the numbers, you are looking to find the number of ways you can roll 5 dice. You need to permute the $E_1$ and $E_2$ events. If 6 can be in two locations, then you have $1^2$ ways of placing 6 in 2 locations. Next, you need to permute the multiset:
$$\{E_1\cdot 2, E_2 \cdot 1, 6\cdot 2\}$$
There are
$$\dfrac{5!}{2!1!2!}$$
ways to permute this multiset.
So, the answer is:
$$\dfrac{3^2\cdot 2^1\cdot 1^2 \cdot \dfrac{5!}{2!1!2!}}{6^5} = \dfrac{5}{72}$$
A: 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 $1$.
A: There are six possible outcomes for each of the five throws, so your denominator should be $6^5$.
For the favorable cases, we must have a sequence of five throws in which three of the outcomes are 1, 2, or 3, two of the outcomes are 4 or 5, and two of the outcomes are a six.  There are $\binom{5}{2}$ ways to select which two positions in the sequence are occupied by a 1, 2, or 3 and $3$ ways to fill each of those positions.  That leaves three positions to fill.  There are $\binom{3}{1}$ ways to choose which of these positions will be occupied by a 4 or 5 and two ways to fill that position.  The remaining two positions of the sequence must be filled with a 6.  Therefore, the number of favorable cases is 
$$\binom{5}{2}3^2 \cdot \binom{3}{1}2^1 \cdot \binom{2}{2}1^2 = \binom{5}{2}\binom{3}{1}\binom{2}{2} \cdot 3^2 \cdot 2^1 \cdot 1^2$$
Hence, the probability of obtaining a 1, 2, or 3 exactly twice and a 4 or 5 exactly once in five rolls of a fair die is
$$\frac{\dbinom{5}{2}\dbinom{3}{1}\dbinom{2}{2} \cdot 3^2 \cdot 2^1 \cdot 1^2}{6^5}$$ 
It is illustrative to compare this result with those provided by drhab and InterstellarProbe.
Observe that 
$$\binom{5}{2}\binom{3}{1}\binom{2}{2} = \frac{5!}{2!3!} \cdot \frac{3!}{1!2!} \cdot \frac{2!}{2!0!} = \frac{5!}{2!1!2!} = \binom{5}{2, 1, 2}$$
where the multinomial coefficient 
$$\binom{5}{2,1,2}$$ 
counts the number of ways of choosing a sequence of length five composed of two objects of one type (in this case, a roll that produces a 1, 2, or 3), one object of a second type (in this case, a roll that produces a 4 or 5), and two objects of a third type (in this case, a roll that produces a 6).
Therefore,
\begin{align*}
\frac{\dbinom{5}{2}\dbinom{3}{2}\dbinom{2}{2} \cdot 3^2 \cdot 2^1 \cdot 1^2}{6^5} & = \frac{\dfrac{5!}{2!1!2!} \cdot 3^2 \cdot 2^1 \cdot 1^2}{6^5}\\
& = \frac{5!}{2!1!2!}\left(\frac{3}{6}\right)^2\left(\frac{2}{6}\right)^1\left(\frac{1}{6}\right)^2\\
& = \binom{5}{2,1,2}\left(\frac{3}{6}\right)^2\left(\frac{2}{6}\right)^1\left(\frac{1}{6}\right)^2
\end{align*}
