How many ways can we place $3$ chess pieces so that none are in the same column or row? Chess pieces are distinguishable, so we can imagine them as one pawn, one knight, one rook. A chess board is a $8\times 8$ grid.
So the correct answer is $8^27^26^2$ because we have $8$ choices for the column and $8$ for the row to place the first piece, and $7$ choices for the second piece, and so on.
However, when I attempted the question, I thought that we could do $(8 C 3)\cdot (8 C 3)$, which first counts the number of ways to choose $3$ different columns and then $3$ different rows. However that answer seems to undercount. So why is it wrong and why is it undercounting?