combinatorics
If I have a $4\times 4$ chess board, in how many ways can I place four distinct pawns on the board such that each column and row of the board contains no more than one pawn?
I have a feeling this is a basic combination problem with cases. I don't see it yet.
For bishops choose a colour - in one direction you will find seven diagonals. So at most one on each diagonal, two colours, fourteen bishops - realised by eight on the first rank and six on the eighth.
For rooks, one on each rank - eight along the diagonal will do.
For knights - thirty two all on the same colour.
Not sure what you mean about pawns.
This on a standard chessboard, but generalisable. If the side of the chessboard is odd, put the knights on the colour with most squares.
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.
Best Answer
We have 4 choices for the column of the pawn in the first row, 3 choices for the column of the pawn in the second row, 2 choices for the column of the pawn in row 3, and 1 choice for the column of the pawn in row 4, for a total of $4!$ places for the positions of the pawns.
Since the pawns are distinct, there are $4!$ ways to place them in these chosen positions; so there are $4!\cdot4!=576$ possibilities.
Alternatively, there are 16 places for the first pawn, then 9 places for the second pawn, only 4 choices left for the third pawn, and just 1 choice for the fourth pawn, giving $\;16\cdot9\cdot4\cdot1=576$ possibilities.