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.


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?

  • 3
    $\begingroup$ yes $ $ $ $ $ $ $\endgroup$ – Exodd Nov 30 '19 at 18:44

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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