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I am interested how to calculate the number of arrangements of a given set of chess pieces on a given board that can be of a non-square rectangular shape too.

( I am not a native speaker so please correct me in case I used any word here incorrectly )

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Suppose you have an $m \times n$ rectangle, and piece types $t_1, \ldots t_N$, and quantities $q_1 \ldots q_N$ of each piece.

If $N=1$, then the answer is simply $\binom{mn}{q_1}$, and by "conditioning on where types $t_1, \ldots t_{N-1}$ went" we get the general answer is $$\prod_{i=1}^N \binom{ nm - \sum_{j < i} q_j}{q_i}$$

More directly (and independent of the choice of ordering!), this can be seen as $$\frac{(mn)(mn-1) \ldots (mn - {\sum_i q_i}+1)}{\prod_i q_i!}$$where we divide out by the product of the factorials to account for the fact that pieces of the same type are indistinguishable.

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yeah in fact including the empty spaces is much nicer: if you include that, the answer is just the multinomial coefficient $$\binom{ mn }{q_1, \ldots q_n,Q}$$where $Q$ is the number of empty spaces. –  uncookedfalcon Jan 7 '13 at 6:38
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