2007
Problem - 1680
A set of $25$ square blocks is arranged into a $5 \times 5$ square. How many different combinations of $3$ blocks can be selected from that set so that no two are in the same row or column?
There are $25$ ways to select the first square. Then there are $16$ ways to select the second square, and $9$ ways to select the third. Meanwhile, the order of choosing these three squares is not important. Therefore, the duplicate factor is $3!$. Hence, the final answer is $$\frac{25\times 16\times 9}{3!}=\boxed{600}$$