You roll 5 dice. Find the probability of 2 pairs. I have seen this answered on here before. I have a slightly different form of answer, also my answer differs from the solution given.
My answer is
$${6 \choose 3} {5 \choose 2,2,1}6^{-5}.$$
${6 \choose 3}$ comes from choosing 3 faces out of 6 i.e $(2,3,1)$. We then arrange 5 symbols two pairs of which are identical. i.e $(2,2,3,3,1), (1,1,3,2,2), (3,3,2,1,1)$. This gives us the term ${5 \choose 2,2,1}$.
The answer provided in the book gives me
$$4{6 \choose 2} {5 \choose 2,2,1}6^{-5}.$$
Since $ {5 \choose 2,2,1} = {5 \choose 4}{4 \choose 2}$ this can be written
$$4{6 \choose 2} {5 \choose 4}{4 \choose 2}6^{-5}.$$
I just want to understand where I am going wrong. Also, I would like to know how to get
$$4{6 \choose 2} {5 \choose 4}{4 \choose 2}6^{-5}.$$
from first principles.
EDIT
The following
$${6 \choose 3} {5 \choose 2,2,1}6^{-5},$$
is incorrect. Because   ${6 \choose 3} $ would "include" only $\{11223\}$ instead of both  $\{22331\}$ and  $\{11223\}$
 A: Expanding on @user51547's comment, after choosing the 3 faces and grouping of the 5 dice, you also need to decide the assignment of the faces to the dice. In this case, we just need to decide which face is the unique one (not paired), and the remaining two must be the pair faces. So there is an extra $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ choice, meaning your answer becomes
$$\begin{pmatrix} 6 \\ 3 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} \begin{pmatrix} 5 \\ 2,2,1 \end{pmatrix} 6^{-5}$$
If you expand the first two factors and rearrange, you get the form that book provided you.
For your second question on getting from first principles, I'll first make a slight modification
$$4 \begin{pmatrix} 6 \\ 2 \end{pmatrix} \begin{pmatrix} 5 \\ 4 \end{pmatrix} \begin{pmatrix} 4 \\ 2 \end{pmatrix} 6^{-5} = \begin{pmatrix} 6 \\ 2 \end{pmatrix} \begin{pmatrix} 4 \\ 1 \end{pmatrix} \begin{pmatrix} 5 \\ 4 \end{pmatrix} \begin{pmatrix} 4 \\ 2 \end{pmatrix} 6^{-5}$$
Now it can be explained as

*

*Among the 6 possible faces, choose the 2 pairing faces

*Among the remaining 4, choose the 1 unique face

*Among the 5 dice, choose the 4 to have the pairing faces

*Among the 4 pairing dice, choose which 2 have the first pairing face

Note that here I am assigning which faces are pairing/non-pairing while I am choosing them.
