A probability question that involves $5$ dice For five dice that are thrown, I am struggling to find the probability of one number showing exactly three times and a second number showing twice.
For the one number showing exactly three times, the probability is:
$$
{5 \choose 3} \times \left ( \frac{1}{6} \right )^{3} \times \left ( \frac{5}{6}\right )^{2}
$$
However, I understand I cannot just multiply this by 
$$
{5 \choose 2} \times \left ( \frac{1}{6} \right )^{2} \times \left ( \frac{5}{6}\right )^{3}
$$
as this includes the probability of picking the original number twice which allows the possibility of the same number being shown $5$ times. I am unsure of what to do next, I tried to write down all the combinations manually and got $10$ possible outcomes so for example if a was the value found $3$ times and $b$ was the value obtained $2$ times one arrangement would be '$aaabb$'. However I still am unsure of what to do after I get $10$ different possibilities and I am not sure how I could even get the $10$ different combinations mathematically. Any hints or advice on what to do next would be much appreciated. 

First, I assume they wil all come out in neat order, first three in a row, then two in a row of a different number. The probability of that happening is 
$$
\frac{1}{6^2}\cdot \frac{5}{6}\cdot\frac{1}{6} = \frac{5}{6^4}
$$
The first die can be anything, but the next two have to be equal to that, so the $\frac{1}{6^2}$ comes from there. Then the fourth die has to be different, and the odds of that happening is the $\frac{5}{6}$ above, and lastly, the last die has to be the same as the fourth.
Now, we assumed that the three equal dice would come out first. There are other orders, a total of $\binom{5}{3}$. Multiply them, and you get the final answer
$$
\binom{5}{3}\frac{5}{6^4}
$$

You might reason another way. Let the event $A$ be "there are exactly 3 of one number" and $B$ be "There are exactly 2 of one number". Then we have that 
$$
P(A\cap B) = P(A) \cdot P(B|A) = {5 \choose 3} \cdot \frac{1}{6^3} \cdot\frac{5^2}{6^2} \cdot \frac{1}{5} = \binom{5}{3}\frac{5}{6^4}
$$
Where $P(A)$ you allready calculated on your own, and $P(B|A)$ is the probability that there is an exact pair given that there is an exact triple, which again is the probability that the two last dice are equal, and that is $\frac{1}{5}$.
