Choosing squares from a grid so that no two chosen squares are in the same row or column 
How many ways can 3 squares be chosen from a 5x5 grid so that no two
  chosen squares are in the same row or column?

Why is this not simply $\binom{5}{3}\cdot\binom{5}{3}$?
I figured that there were $\binom{5}{3}$ ways to choose $3$ different "$x$-axis" coordinates and then same for the "$y$-axis". Thus I would multiply them.
Thanks

You can choose the first square arbitrarily, so there are $25$ options for it. 
There are now $4$ rows and $4$ columns that the second square can be chosen from, so there are $16$ options for it (alternatively, if one just wants to count, there are $25$ squares total, minus the chosen first square, minus the $4$ other squares in the same row, minus the other $4$ squares in the same column, for $25 - 1 - 4 - 4  = 16$ options). 
Similarly, there are $9$ options for the third square. 
Dividing by $3!$ to account for the fact that the same three squares can be chosen in that many ways,  the correct answer is
$$\frac{25\cdot 16\cdot 9}{3!} = 600$$
not
$$\binom{5}{3}\cdot\binom{5}{3}=10\cdot 10=100.$$
