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?

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*}
