Probability problem regarding rooks on a chessboard

Question

Eight rooks are placed in distinct squares of an 8 x 8 chessboard, with all possible replacements being equally likely. Find the probability that all the rooks are safe from one another.

Answer 1

Since rooks attack horizontally and vertically, you can't have a rook in the same row or column as another.

So in the first row, you place a rook. There are 8 possible places. In the next row you place a rook. It can't be in the same column as the other rook, so 7 possible places. So on and so on, to get $8!$ possible ways to place the rooks without them threatening one another.

That's your desired solution. You need to divide by all possible ways to place the rooks.

There are 64 spaces on a chess board and you need to place rooks in 8 of them. So it's just $64\choose8$.

That makes your answer $$\frac{8!}{64\choose8}$$

Answer 2

Thought process: - First rook can be placed in 64 squares.

- For the next rook the entire row and column of the first rook are off-limits, so the second rook has 64-16+1=49 squares available (add one to avoid double counting the square where the first rook was placed).

- repeat the same idea starting with 49 and add back double counted squares from the rows and columns of previous rooks already placed.

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Calculating Probablity:

A:= placing 8 rooks without either of them capturing each other.

A_i:= placing the ith rook

first rook can be placed in 64 squares

second rook can be placed in 64-16+1= 49 squares

third rook can placed in 49-16+3= 36 squares

fourth rook can placed in 36-16+5= 25 squares

fifth rook can placed in 25-16+7= 16 squares

sixth rook can placed in 16-16+9= 9 squares

seventh rook can placed in 9-16+11= 4 squares

eighth rook can placed in 4-16+13= 1 square

By principle of counting |A| = |A_1|*|A_2|*|A_3|*|A_4|*|A_5|*|A_6|*|A_7|*|A_8| = 64*49*36*25*16*9*4*1 outcomes

P(A) = |A| / |S| = 64*49*36*25*16*9*4*1 / 64*63*62*61*60*59*58*57