On a chess board one white square is chosen at random. In how many ways can a black square be chosen such that it does not lie in the same row as the white square?
thank you!!
2,002
2002? There are only 64 squares on the chess board.
if by "row" you mean in either direction (as a rook can move) then since there are 32 black squares available, and 8 of them are in the same row/column as the white square chosen, that leaves 24 available black squares to choose from.
Yes, but combine them all and all the different possible moves, that concludes it to 2,002
To solve this problem, let's consider the chessboard and mark the selected white square.
We can choose a black square in two ways:
1. If the white square is in an even row (e.g., row 2, row 4), then there are four rows available for the black square (rows 1, 3, 5, and 7). So, there are 4 possibilities.
2. If the white square is in an odd row (e.g., row 1, row 3), then there are three rows available for the black square (rows 2, 4, and 6). So, there are 3 possibilities.
Since we have an equal number of even and odd rows on a chessboard, we can conclude that there are an equal number of white squares in even and odd rows.
Therefore, the total number of ways to choose a black square that does not lie in the same row as the white square is (number of ways for even rows) + (number of ways for odd rows) = 4 + 3 = 7.
Hence, there are 7 ways to choose a black square that does not lie in the same row as the white square.