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

Here's the hard way to do the problem: inclusion-exclusion. 
There are $25\choose3$ ways to choose 3 squares from the 25. 
Now you have to subtract the ways that have two squares in the same row or column. There are 10 ways to choose the row/column, $5\choose2$ ways to choose the two squares in the row/column, and 23 choices remaining for the third square, so all up you must subtract $10\times{5\choose2}\times23$. 
Now you have to put back in all the configurations you subtracted out twice. These are the ones with two in the same row and two in the same column, of which there are $25\times4\times4$, and also the ones in which there are three in the same row/column, of which there are $10\times{5\choose3}$. The ones with 3 in a column were counted in once, then subtracted out 3 times, so they have to be put back in twice.
So the answer is $${25\choose3}-10\times{5\choose2}\times23+25\times4\times4+2\times10\times{5\choose3}$$ which comes out to 600. 
