How to find the amount of possible piece configurations in the game of Checkers/Draughts? All I know is that factorials are involved, I hope someone here can help me out.
These are the restrictions:


*

*32 habitable squares

*Each square can be in one of five states: [white man, black man, white king, black king, empty]

*12 is the maximum amount of any color of pieces (white men + white kings <= 12)


The calculation doesn't have to be absolutely perfect, symmetries can be ignored, I also want to figure out different kinds of combinations (for example given how long a game lasts on average how many different board positions one can expect etc...) so a general explanation of the formula/reasoning to be used would be preferred to the direct answer, although both will be appreciated.
 A: Acknowledgement: my methodology was inspired by quasi's answer.
In a game of checkers/draughts non-king pieces cannot appear on the last row because they immediately become kings. This gives these white pieces and black pieces partially overlapping areas consisting of 4 squares in their own first row plus 24 shared squares.
Define:


*

*$w_1$: white pieces in white's first row.

*$b_1$: black pieces in black's first row.

*$w_m$: white pieces in the shared middle six rows.

*$b_m$: black pieces in the shared middle six rows.

*$w_k$: white kings, can be placed anywhere.

*$b_k$: black kings, can be placed anywhere.


Then the number of valid positions is given by:
$$\left\{\sum_{w_1=0}^4\sum_{b_1=0}^4\sum_{w_m=0}^{12-w_1}\sum_{b_m=0}^{12-b_1}\sum_{w_k=0}^{12-w_1-w_m}\sum_{b_k=0}^{12-b_1-b_m}\binom{4}{w_1}\binom{4}{b_1}\binom{24}{w_m}\binom{24-w_m}{b_m}\\\binom{32-w_1-b_1-w_m-b_m}{w_k}\binom{32-w_1-b_1-w_m-b_m-w_k}{b_k}\right\}-1$$
The $-1$ is needed because an empty board is not a valid position.
This evaluates to $500,995,484,682,338,672,639$.
Between 1989 and 2007 the Chinook draughts player was developed. Linked on that Wikipedia page is a paper Checkers is Solved which confirms the above number.
A: Let $w,b$ be integers with $0 \le w \le 12$ and $0 \le b \le 12$.

Then for configurations having exactly $w$ white pieces and $b$ black pieces, the number of combinatorially legal configurations (even if not reachable by the rules of the game) is
$$\binom{32}{w}\binom{32-w}{b}2^{w+b}$$
Explanation:


*

*$\large{\binom{32}{w}}$ is the number of ways to place $w$ white pieces on an empty board.$\\[14pt]$

*$\large{\binom{32-w}{b}}$ is the number of ways to place $b$ black pieces on the board, given that $w$ positions are already occupied by white pieces.$\\[8pt]$

*Each placed piece can be either a King or a non-King. Thus, for each piece,  there are two choices, so we get a factor of $2^{w+b}$.


Hence, the total number of combinatorially legal configurations, for all possible values of $w,b$ is
$$\sum_{b=0}^{12}\sum_{w=0}^{12} \binom{32}{w}\binom{32-w}{b}2^{w+b}$$
which equals $2,308,487,701,529,742,077,825$ or approximately $2.3\times10^{21}$.

Of course, this is just an upper bound, since some of the above configurations, though combinatorially legal, can't actually be achieved by a sequence of legal moves, based on the standard starting positions.

As regards your question about the length of an "average" game, if two players have fixed strategies, all games they play will have the same length (and the same outcome), so in that context, average length is not very meaningful.

On the other hand, for the sake of exploration, suppose both players make random moves, consistent with the rules of the game (i.e., if jumps are forced, they must make such a move). Assume also that a player always converts a piece to King whenever it reaches the last rank. Now the concept of average length is meaningful, but the exact average, while it exists theoretically, is surely intractable to actually compute. Instead, write a program to implement such random games, and then run a simulation, to get an approximate value for the average game length.
