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In how many ways can you place 32 chess pieces on a standard (8x8) chessboard? This does not have to comply with the rules of the game.

My answer is:

We place the pieces as though they all are distinguishable, i.e.

$\frac{64!}{32!}$

but we cannot distinguish between eight white pawns, two white knights, two white rooks, two white bishops and the same with the black ones. So the final answer is

$\frac{64!}{32!}\cdot\frac{1}{{8!}^2 2!^22!^22!^2}$

Is my reasoning correct?

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    $\begingroup$ yes $ $ $ $ $ $ $\endgroup$
    – Exodd
    Nov 30, 2019 at 18:44

1 Answer 1

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Since you have $32$ chess pieces and $64$ $(8\times8)$ squares, you first need to choose $32$ squares to use.

This can be done in ${64 \choose 32} = \frac{64!}{32!\times32!}$ ways. Now, for each of these ways, you can have $32!$ ways of arranging the pieces (if all pieces were different). So you have to multiply ${64\choose32}$ with $32!$.

But all pieces aren't different, as you mentioned there are $8$ pawns and hence $8!$ ways of arranging them. So, you were over counting initially - by a factor of $8!$, because all those $8!$ arrangements were essentially the same.

This reasoning would apply to all pieces that are identical and hence you would have to divide ${64 \choose 32} * 32!$ by $(8!\times2!\times2!\times2!)^2$.

Final answer: $\frac{64!}{32!\times(8!\times2!\times2!\times2!)^2} = 4634726695587809641192045982323285670400000$

Fun fact: $1.46966\times10^{35}$ years would contain these many seconds.

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