Arranging $6$ distinct rooks on a $6\times 6$ board 
How many ways there are to arrange $6$ distinct rooks on a $6\times 6$ board such that no rook would threaten another?

My attempt:
1st rook: $36$ options.
2nd rook: $36-10+2=28$, there are 10 squares that the first rook threaten, 2 joint squares so to avoid counting them twice we add them.
3rd rook: $28-10+2=20$, 4th: $12$, 5th: $4$, 6th: only one square left.
So the final answer is: $36\cdot28\cdot20\cdot12\cdot4\cdot1$.
 A: Hint: There must be one rook in each row and one in each column.  Relate the (unordered) set of positions of the rooks to permutations of $1,2,\ldots, 6$.
Then think about labelling the rooks.
A: Robert Israel's hint is correct. This is just to expand on his hint and explain the intuition. 
As a rook threatens every other piece in its own row and column, to fit 6 rooks on a 6x6 chessboard, there must be one rook in each row and in each column.  
Method 1
There is a natural bijection between an (unordered) solution and the permutations on $[1...6]$ defined by the vertical position of the rook in the first column through the sixth column. Thus there are $6!$ (unordered solutions). Now, for each solution there are six spots on the board to populate with 6 distinct rooks, therefore there are $6!$ distinct solutions for each unordered solution. In total there are $(6!)^2$ solutions.
Method 2
The first rook can choose from 6 columns and 6 rows within that column (36 spots). The second rook can choose from the 5 remaining columns and 5 remaining rows (25). The third rook can choose from the 4 remaining columns and 4 remaining rows (16), and so on. Therefore there are $6^2\times5^2\times4^2\times3^2\times2^2\times1^2=(6!)^2$ solutions
