On an $6 \times 6$ chessboard, we randomly place counters on three different squares. What is the probability that no two counters are in the same row or column?
The number of ways to pick three squares is $\left(\begin{array}{c}36 \\ 3\end{array}\right)$.

We now count the number of ways to pick three squares with no two in the same row or column. One can select three distinct rows, in $\left(\begin{array}{l}6 \\ 3\end{array}\right)$ ways, that we will place the counters in. Afterwards, there are 6 ways to pick the column for the first row, 5 to pick the column in the second row, and 4 for the final row.

So the answer is

$$
\frac{\left(\begin{array}{l}
6 \\
3
\end{array}\right) \cdot 6 \cdot 5 \cdot 4}{\left(\begin{array}{c}
36 \\
3
\end{array}\right)}=\frac{20 \cdot 6 \cdot 5 \cdot 4}{36 \cdot 35 \cdot 34 / 6}=\frac{40}{119}
$$