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? 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.
 A: 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.
A: Note: I assumed the pawns, characteristically the weakest and most numerous of chess pieces, are interchangeable (not individually distinguishable), but as user84413 interprets "four distinct pawns", they are distinguishable.  That leads to an extra factor of permutations (of the individual pieces), and hence to a different answer, $(4!)^2 = 24^2 = 576$.

Yes, it is a basic combinatorial problem, with $4!= 24$ solutions.
The key to seeing this is to consider the pawns must be one in each column and one in each row.  Therefore if we enumerate the pawns by row, their column indexes are a permutation of the row indexes.  Hence (for $4$ pawns on a $4\times 4$ board) we get $4!$ placements.
