How many ways we can place $3$ rooks on a $5\times 5$ board such that no two rooks attack each other?
Now, I know that the answer is $600$ but could someone tell me why this method doesn't work,
First there are $25$ way I can choose the first rook, and since placing the fist rook will get rid of $9$ squares on the board (The squares that are in the same row or column as the first rook) so we there are $25-9=16$ ways we can place the second rook. Now we are left with $25-9\cdot 2+2=9$ squares to place the third rook; The $25-9\cdot2$ part is obvious and we add $2$ since there would be a $2$ square intersection between the rows and columns of the first and the second row.
Therefore the answer is $25\cdot 16\cdot 9=3600$? So where is the mistake?
Best Answer
You are treating the rooks as distinguishable. Since they are indistinguishable, you have overcounted by a factor of $3!$, so the answer is $25\cdot 16\cdot 9/3!=600$.