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?
Best Answer
EDIT:
Adding your method and completing it which works too -
You first choose $3$ rows and $3$ columns so number of ways = $(^8C_3)^2$.
This gives you $9$ squares. Please see one of the cases in the picture. Now you can place the first piece on any of the $9$ squares. That leaves $4$ choices for the next piece. The last one has just one places to go to.
So total number of ways = $(^8C_3)^2 \times 9 \times 4 = 112896$.
May be an easier way to look at it is
Number of ways $ = 8 \times 8 + 7 \times 7 + 6 \times 6$. Explanation -
There are $64$ squares on a chessboard. The first one can be placed anywhere out of $64$ squares - (say, Col1 and Row1). Once that is done, the second one has to avoid squares in col1 and row1. That leaves $64-15 = 49 = 7 \times 7$ squares for the second.
Now think of the third one - $2$ squares to avoid will be common between first and second. So add another $13$ squares to avoid. That leaves $36 = 6 \times 6$ squares.