<s>[INST] 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?
 [/INST] ${64 \choose 5}$ looks OK for the denominator
For distinct rows, there are ${8 \choose 5}$ ways of choosing the five rows.

*

*The dot in the first occupied row can be in any of the $8$ columns

*The dot in the second occupied row can be in any of the $7$ remaining columns

*The dot in the third occupied row can be in any of the $6$ remaining columns

*The dot in the fourth occupied row can be in any of the $5$ remaining columns

*The dot in the fifth occupied row can be in any of the $4$ remaining columns

So I would get a probability of $$\dfrac{{8 \choose 5}\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}{{64 \choose 5}}$$
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