"Let's play mini-Sudoku!
We wish to place an "X" in four boxes, such that there is exactly one "X" in each row, column, and 2×2 outlined box. In how many ways can we do this?"
Solution:
"
Using constructive counting and combinations together, we can find 16 possibilities. This is how we find the answer: First, we determine how many possibilities there could be without any restrictions. We do this by choosing one possibility (as seen in the example) and calculating the number of spots you could place the "X" in the grid. In this circumstance -with 16 available slots- there are 384 possible combinations with over-counting. We have drastically over-counted so we divide by the number of X's in one solution factorial (24). When we divide by 24, we get 16 possible ways to solve a mini-Sudoku problem. Let's play!"
Best Answer
I assume the count of $384$ is obtained by considering the four $X$ marks as distinct, and getting actual solutions (obeying row, column, and box restrictions). It's hard to see what "without any restrictions" means, and in fact even with four indistinguishable $X$ there would be $16*15*14*13=43680$ ways to place the $X$ without restrictions, if that means literally anywhere.
It seems it would be harder to get the $384$ count than it would be to assume the $X$ were not distinct to begin with.
One simple solution is as follows. Place two $X$'s, one in each of the upper left and lower right boxes, independently of each other. This gives $4*4=16$ ways to put the $X$'s into those two diagonally opposite boxes. If each of these leads to a unique solution, then $16$ is seen to be the correct count. But once these two $X$ have been placed, each of the remaining boxes (upper right and lower left) has one row and one column in which no $X$ may be placed, leaving exactly one position for an $X$ in that box.