In how many ways can you place 32 chess pieces on a standard (8x8) chessboard?

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.

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?

• yes    – 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.