If I have a checkers board $8\times 8$ and $16$ black pieces , $16$ white ones (this isn't the usual checkers).
Assume that these pieces can be placed anywhere on the board.
How can I calculate the number of ways the board can be laid out?
(it's trivial to say that each position may be a white piece , black piece or empty, but I have an upper bound on the number of checker pieces).
edit:
Considering the upper bound (i.e. $16$ or less black/white piece) how does the number of ways change?
Now, assume that pieces can upgrade to kings. Needless to say, when you gain a king, you lose a normal piece, which preserves the the max 16 pieces rule.
How can I calculate the number of ways the board can be laid out considering the kings?
$${64\choose16}{48\choose16}$$ Multiply by $2^{32}$ to get the answer when there can be kings.
If you have at most 16 black and at most 16 white, you get $$\sum_{m,n=0}^{16}{64\choose m}{48\choose n}$$ which is $$\left(\sum_0^{16}{64\choose m}\right)\left(\sum_0^{16}{48\choose n}\right)$$