Permutation with constrained repetititons

Question

The question is as follows:

How many ways can 12 identical white and 12 identical black pawns be placed on the black squares of an 8 x 8 chessboard

My answer was $\frac{32!}{12!*12!}$

But the correct answers is $\frac{32!}{12!*12!*8!}$ How come?

Answer 1

You can choose $12$ places out of $32$ for the white pawns, then $12$ places out of the remaining $20$ for the black pawns.

$$\binom {32}{12}\cdot \binom {20}{12}=\frac {32!}{12!\cdot 20!}\cdot\frac {20!}{12!\cdot 8!}=\frac {32!}{12!\cdot 12! \cdot 8!}$$

Note that $12+12+8=32$, so this is the formula for splitting the squares between three collections - white pawns, black pawns and empty squares. The formula has to be the same whichever way you count, which explains the three factorials in the denominator. If you wanted twelve black pawns and eight empty squares (the rest being white pawns) you would expect the same result.

Answer 2

First you choose $12$ places out of $32$ for the white: $\binom{32}{12}$ possibilities.

After that you choose $12$ places out of $20$ for the black: $\binom{20}{12}$ possibilities.

That gives a total of $$\binom{32}{12}\binom{20}{12}=\frac{32!}{12!12!8!}$$ possibilities.