- #1
mcafej
- 15
- 0
I'm just checking my work on this. Given an 8x8 chessboard, you randomly place 8 rooks on the board. What is the probability that no rooks can capture another one. In other words, probability that no 2 rooks are in the same row or column.
My solution is simply 8!/(64 choose 8), but that seems like a really small probability. My reasoning behind it is that if you look row by row, in order for 8 rooks to not be able to capture each other, there must only be 1 rook in each row, so if you look at the first row, you have 8 possible choices to put a rook. Then if you look at the next row down, there are only 7 possibilities (because the 8th square on that row would put hat rook in line with the 1st rook), then for the 3rd row there are 6 possible choices, and so on all the way down to 1. So that is all the ways that no two rooks would capture each other, and there are a total of 64 Choose 8 ways to put the rooks on the board.
I'm just double checking my work because I'm studying for a test.
My solution is simply 8!/(64 choose 8), but that seems like a really small probability. My reasoning behind it is that if you look row by row, in order for 8 rooks to not be able to capture each other, there must only be 1 rook in each row, so if you look at the first row, you have 8 possible choices to put a rook. Then if you look at the next row down, there are only 7 possibilities (because the 8th square on that row would put hat rook in line with the 1st rook), then for the 3rd row there are 6 possible choices, and so on all the way down to 1. So that is all the ways that no two rooks would capture each other, and there are a total of 64 Choose 8 ways to put the rooks on the board.
I'm just double checking my work because I'm studying for a test.