Example 5 As shown in Figure 4, place 3 identical coins sequentially into a $4 \times 4$ square grid (each square can only hold 1 coin). Find the probability that any two of the 3 coins are neither in the same row nor in the same column.
Explanation: (1) Calculate the total number of ways to place the coins $n$: The first coin can be placed in 16 different cells, which gives 16 ways; the second coin can be placed in the remaining 15 cells, which gives 15 ways; the third coin can be placed in the remaining 14 cells, which gives 14 ways.
So, the total number of ways to place the coins is
$$
n=16 \times 15 \times 14=3360 \text {. }
$$
(2) Calculate the number of ways to place the coins that meet the problem's requirements $m$: The first coin can be placed in 16 different cells, which gives 16 ways. There are 9 cells that are neither in the same row nor in the same column as the first coin. Therefore, the second coin, which must be in a different row and column from the first coin, has 9 ways to be placed. There are 4 cells that are neither in the same row nor in the same column as the first two coins. The third coin can be placed in the remaining 4 cells, which gives 4 ways.
So, the number of ways to place the coins that meet the problem's requirements is
$$
m=16 \times 9 \times 4=576 \text {. }
$$

Thus, the required probability $P=\frac{m}{n}=\frac{16 \times 9 \times 4}{16 \times 15 \times 14}=\frac{6}{35}$.