Find the probability of placing $5$ dots on an $8 \times 8$ grid. $5$ dots are placed at random on an $8 \times 8$ grid such that no cell has more than $1$ dot.
What is the probability that no row or column has more than $1$ dot?
I thought about this in the following way, the number of ways to place $5$ dots on $64$ cells is $$\begin{pmatrix} 64 \\ 5 \end{pmatrix} $$
Now if I want the $8 \times 8$ to have no row or column with more than $1$ dot then I reasoned that the dots must be placed in the diagonal which has $8$ cells, so the number of ways I can place $5$ dots into $8$ cells is $$\begin{pmatrix} 8 \\ 5 \end{pmatrix} $$
So the desired probability would be $$\frac{ \begin{pmatrix} 8 \\ 5 \end{pmatrix} }{\begin{pmatrix} 64 \\ 5 \end{pmatrix} }$$
However I know that this is not the answer, the correct answer is $$\frac{ \begin{pmatrix} 8 \\ 5 \end{pmatrix} 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4  }{\begin{pmatrix} 64 \\ 5 \end{pmatrix} }$$
but I can't see what the term $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$ counts, can you explain how this term came?

Pick $5$ of $8$ rows to occupy. Sort those in ascending order. Hence the order they were picked in will be forgotten. This gives:
$$
8C5=\binom 85
$$
Pick $5$ of $8$ columns to occupy. Pair those with the ordered set of chosen rows in the order you pick them. Hence order counts this time, and we have:
$$
8P5=8!/(8-5)!=8\cdot 7\cdot 6\cdot 5\cdot 4
$$
which is where the extra product comes from.
